D.C. Circuit Numericals

1. D.C. Circuit (Numerical) 2. Emf of cell. 3. Heating effect

Type: 1 (Current, Drift velocity, Resistance and Resistivity)

Formula:

1. Electric current (I) = $\frac{\text{charge (q)}}{\text{time (t)}}$

2. $I = V_d e n A$

3. Current density, $J = \frac{I}{A}$ i.e. $J = V_d e n$

$4. V = IR$

5. $R = \rho \frac{l}{A}$

Q. 1. A copper wire has a diameter of 1.02mm and carries a constant current of 1.67A. If the density of free electrons in copper is $8.5 \times 10^{28}$ m^-3, calculate the current density and drift velocity of the electrons.

Here,

Diameter, $d= 1.02mm = 1.02 \times 10^{-3}m$

Current, $I = 1.67A$

Density of free electrons, $n = 8.5 \times 10^{28} \text{m}^{-3}$

(i) Current density, $J = ?$

(ii) Drift velocity, $V_d = ?$

We have,

$I = V_d e n A$

(ii) $\begin{aligned} V_d = & \frac{I}{enA} = \frac{I}{en \left[\pi d^2/4\right]} \\ = & \frac{1.67}{1.6 \times 10^{-19} \times 8.5 \times 10^{28}} \times \frac{4}{\pi (\ 1.02 \times 10^{-3})^2} \\ = & 1.5 \times 10^{-4} \text{ms}^{-1} \end{aligned}$

(i) Current density, $J = V_d e n$

or, $J = 1.5 \times 10^{-4} \times 1.6 \times 10^{-19} \times 8.5 \times 10^{28}$ $= 2.04 \times 10^{6} \text{ Am}^{-2}$

(i) Alternative method

$J = \frac{I}{A} = \frac{I}{\pi d^2/4}$

$= \frac{1.67 \times 4}{\pi (1.02 \times 10^{-3})^2} = 2.04 \times 10^6 \text{ Am}^{-2}$

Q. 2.Resistance of a wire of length 1m, diameter 1mm is 2.2 $\Omega$ . Calculate its resistivity and conductivity. Ans: $1.728\times10^{-6}\Omega$ m, $579038.8\Omega^{-1}$ m^-1

Here,

Length, $l = 1$ m

Diameter, $d = 1mm = 1 \times 10^{-3}m$

Resistance, $R = 2.2\Omega$

• (i) Resistivity, $\rho = ?$
• (ii) Conductivity, C = ?

We have,

$R = \rho \frac{l}{A}$

or, $\rho = \frac{RA}{I} = \frac{R}{I} \times \frac{\pi d^2}{4}$

$=\,\frac{2.2\times\pi(1\times10^{-3}\,)^2}{1\times4}=1.728\times10^{-6}\Omega\,m$

(ii) Conductivity, $C = \frac{1}{\text{resistivity}} = \frac{1}{\rho}$

$= \frac{1}{1.728 \times 10^{-6}} = 578745.2\Omega^{-1} \text{m}^{-1}$

Q. 3. A potential difference of 4.5V is applied between the ends of wire that is 2.5m long and has radius of 0.654mm. The resulting current through the wire is 17.6A. What is the resistivity of the wire?

Ans: $1.376 \times 10^{-7} \Omega m$ .

Here,

Potential difference, V = 4.5 V

Length of the wire, l = 2.5m

Radius of the wire, $r = 0.654 \text{mm} = 0.654 \times 10^{-3} \text{m}$

Current, I = 17.6A

Resistivity, $\rho = ?$

We have,

Resistance, $R = \rho \frac{l}{A}$

or, $\rho = \frac{RA}{l}$

or, $\rho = \pi r^2 \frac{R}{l}$ …(i)

From Ohm’s law,

$V = IR$

or, $R = \frac{V}{I}$

From eqn (i),

$\rho = \pi r^2 \, \frac{V}{I \times \mathit{l}}$

or, $\rho = \pi \times (0.654 \times 10^{-3})^2 \times \frac{4.5}{17.6 \times 2.5}$ $= 1.374 \times 10^{-7} \Omega m$

Q. 4. A copper wire has a diameter of 1.02mm, cross-sectional area $8.2 \times 10^{-7} \text{m}^2$ and resistivity $1.72 \times 10^{-8}$ $\Omega$ m. It carries a current of 1.67A. Find the electric field magnitude in the wire and the potential difference between two points in the wire 50m apart.

Here,

Diameter of the wire, $d = 1.02 \text{mm} = 1.02 \times 10^{-3} \text{m}$

Area of cross-section, $A = 8.2 \times 10^{-7} \text{m}^2$

Resistivity, $\rho = 1.72 \times 10^{-8} \Omega m$

Current, I = 1.67A

(ii) P.d., $V = ?$

Distance between two points (i.e. length), l = 50m

We have,

(ii) From Ohm’s law,

$V = IR$

Since, Resistance, $R = \rho \frac{l}{A}$

: $V = I \times \rho \frac{l}{A} = \frac{1.67 \times 1.72 \times 10^{-8} \times 50}{8.2 \times 10^{-7}}$

or, P.d, $V = 1.75V$

(i) Electric field, $E = \frac{V}{l} = \frac{1.75}{50}$

= 0.035 Vm^-1

Q. 5. A tightly coiled spring having 75 coils each 3.50 cm in diameter, is made of insulated metal wire 3.25mm in diameter. An ohm meter connected across its opposite ends reads $1.74\Omega$ . What is the resistivity of the metal?

Ans: $1.75\times10^{-6}\Omega$ m

Here,

No. of turns, n = 75

Diameter of coil, $D = 3.5 \text{cm} = 3.5 \times 10^{-2} \text{m}$

Diameter of the wire, $d = 3.25 \text{mm} = 3.25 \times 10^{-3} \text{m}$

Resistance, $R = 1.74\Omega$

Resistivity, $\rho = ?$

We have,

Resistance, $R = \rho \frac{l}{A}$

$\therefore \text{Resistivity}, \quad \rho = \frac{\text{RA}}{l}$

or, $\rho = \frac{R}{nC} \times \frac{\pi d^2}{4}$ here, $l = nC$ , C is circumference of coil

$\text{or,} \qquad \rho = \frac{R}{n \times \pi D} \times \frac{\pi d^2}{4} = \frac{1.74 \times (\ 3.25 \times 10^{-3})^2}{75 \times 3.5 \times 10^{-2} \times 4} = 1.75 \times 10^{-6} \Omega m$

Type: 2

(Variation of Resistance with Temperature)

Formula:

Variation of Resistance with Temperature,

$R_{\theta} = R_0 (1+\alpha\theta)$ , $\alpha$ is Temperature coefficient of resistance.

Q. 6. The resistance of a conductor is $10\Omega$ at $50^{\circ}$ C and $15\Omega$ at $100^{\circ}$ C. Calculate its resistance at $0^{\circ}$ C.

Given,

At temperature, $\theta_1 = 50^{\circ}$ C, Resistance, $R_{\theta_1} = 10\Omega$

At temperature, $\theta_2 = 100^{\circ}$ C, Resistance, $R_{\theta 2} = 15\Omega$

At temperature, $\theta = 0^{\circ}$ C, Resistance, $R_0 = ?$

From the variation of resistance with temperature, we have,

$R_{\theta} = R_0 (1+\alpha\theta),$

For 50°C, $R_{\theta 1} = R_0 (1+\alpha\theta_1)$

$10 = R_0 (1+\alpha \times 50)....(1)$

For $100^{\circ}$ C, $R_{\theta 2} = R_0 (1 + \alpha \theta_2)$

$15 = R_0 (1+\alpha \times 100)...(2)$

Dividing (2) by (1)

$1.5 = \frac{1 + 100\alpha}{1 + 50\alpha}$

or, $1.5 + 75\alpha = 1 + 100\alpha$

or, $25\alpha = 0.5$

or, $\alpha = \frac{0.5}{225} = 0.02 \, {}^{\circ}\text{C}^{-1}$

From (1), $R_0 = \frac{10}{1+50\times0.02} = \frac{5\Omega}{1+50\times0.02}$

Type: 3

(Series and Parallel combination)

Formula:

In Series Combination:

• (i) Current is same.
• (ii) Voltage drop is different.
• (iii) $V_1 = IR_1$ , $V_2 = IR_2$ , $V_3 = IR$
• (iv) Total potential (V) = $V_1 + V_2 + V_3$
• (v) $R_s = R_1 + R_2 + R_3$

Parallel Combination:

• (i) Current is different.
• (ii) Voltage drop is same, $V_1 = V_2 = V_3$
• (iii) $V_1 = I_1R_1 \ V_2 = I_2R_2 \ V_3 = I_3R_3$ ,
• (iv) $I = I_1 + I_2 + I_3$
• (v) $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
• (vi) If there are only two resistors in parallel then to find equivalent resistance, we use the formula

$R_{eq} \text{ or } R_p = \frac{R_1 \times R_2}{R_1 + R_2} \text{ instead of } \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

Q.7. What is the potential difference across $100\Omega$ resister in the circuit given below?

Given,

Emf, E = V = 12V

P.d. across $100\Omega$ , $V_1 = ?$

$R_{\rm AB} = \frac{100{\times}50}{100 + 50} = \frac{100}{3}$

$R_{eq} = \frac{100}{3} \, + 500 = \frac{1600}{3}$

Now, $E = I R_{eq}$

$I = \frac{E}{R_{eq}} = \frac{12}{1600/3} = 0.0225A$

Then, P.d. across $500\Omega = 0.0225 \times 500$ [V= IR]
= 11.25 V

∴ P.d. across $100\Omega$ (AB) = $12-11.25 = 0.75 \text{ V} = \frac{3}{4} \text{ V}$

Q.8. Alternative Method

Given,

Emf, $E = V = 12V$

P.d. across $100\Omega$ , $V_{I} = ?$

$R_{AB} = \frac{100 \times 50}{100 + 50} = \frac{100}{3}$

$R_{eq} = \frac{100}{3} + 500 = \frac{1600}{3}$

Now, $E = I R_{eq}$

$I = \frac{E}{R_{eq}} = \frac{12}{1600/3} = 0.0225A$

Since, $R_1$ & $R_2$ are in parallel

$\therefore V_1 = V_2$

$I_1R_1=I_2R_2$

$I_1 \times 100 = (I - I_1) \times 50$

$100I_1 = 50I - 50I_1$

$150I_1 = 50I$

$I_1 = \frac{50{\times}0.0225}{150}$

So, P.d across $100\Omega$ , $V_1 = I_1R_1$

$=\frac{50\times0.0225}{150}\times100$

$=\frac{3}{4}$ V = 0.75V

Q.9. Consider the figure below. The current through $6\Omega$ resister is 4A in the direction shown. What are the currents through the $25\Omega$ and $20\Omega$ resistors?

Given,

Current through $6\Omega$ resistor, $(I_1) = 4A$

Current through $25\Omega$ resistor, (I₃) =

Current through $20\Omega$ resistor, (I₄) =

Since, $6\Omega \& 8\Omega$ are in parallel.

∴ Current through $25\Omega$ , $I_3 = I_1 + I_2$

$=4+3=7A$

Again arm AC is parallel to arm XY.

$$\begin{split} \therefore V_{XY} &= V_{AC} \\ I_4 R_4 &= I_3 R_{AC} \\ I_4 R_4 &= I_3 \left( \frac{4 \times 6}{6 + 8} + 25 \right) \\ I_4 &= \frac{I_3}{R_4} \left( \frac{4 \times 6}{6 + 8} + 25 \right) \\ &= \frac{7}{20} \left( \frac{4 \times 6}{6 + 8} + 25 \right) \\ \therefore I_4 &= 9.95 \, \text{A} \end{split}$$

Q.10. In the given figure, the current flowing through the $3\Omega$ resistor is 0.8A. Find the potential drop across $4\Omega$ resistor.

Given, Current through $3\Omega$ resistor, $(I_1) = 0.8A$ Potential drop across $4\Omega$ resistor, $(V_3) = ?$ Since, $3\Omega \& 6\Omega$ are in parallel.

$$\begin{split} \therefore V_1 &= V_2 \\ I_1 R_1 &= I_2 R_2 \\ I_2 &= \frac{I_1 R_1}{R_2} = \frac{0.8 \times 3}{6} = 0.4 \, \text{A} \\ \therefore I &= I_1 + I_2 = 0.8 + 0.4 = 1.2 \, \text{A} \\ \text{Now, P.d across } 4\Omega, \, V_3 &= I R_3 \\ V_3 &= 1.2 \times 4 = 4.8 \, \text{V} \end{split}$$

Q. 11. In the given circuit, calculate the potential difference between the points B and D.

P.d between B & d = ?

$R_{ABC} = 6 + 12 = 18 \Omega$

$R_{ADC} = 12 + 6 = 18 \Omega$

$R_{eq} = \frac{18{\times}18}{18{+}18} = 9\Omega$

Now, $E = I R_{eq}$

$I = \frac{E}{R_{eq}} = \frac{6}{9} = \frac{2}{3} A$

Since, $R_{ABC} = R_{ADC} = 180 \Omega$

So the current will be equally divided

$I_1 = I_2 = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} A$

P.d across AB = $I_1 \times R_{AB} = \frac{1}{3} \times 6 = 2V$

$P.d\ across\ AD = I_1 \times R_{AD} = \frac{1}{3} \times 12 = 4V$

$\therefore$ P.d between B & D = $V_{AD}$ - $V_{AB}$ = 4 -2 = 2V

Emf, Terminal p.d. and internal resistance

Q. 12. A cell of emf 18 V has an internal resistance of 3 $\Omega$ . The terminal p.d. of the battery becomes 15 V when connected by a wire. Find the resistance of the wire.

Here,

Emf of cell, E = 18 V

Internal resistance of cell, $r = 3 \Omega$

Terminal p.d. of the battery, V = 15 V

Resistance of the wire, R = ?

We have circuit formula,

$E = V + Ir$

Or, $I = \frac{E - V}{r}$

Or, $I = \frac{18 - 15}{3} = 1 \text{ A}$

Again, using Ohm’s law,

$V = IR$

Or, $R = \frac{V}{I}$

Or, $R = \frac{15}{1}$

Q.13. In the given figure, when switch s is open, the voltmeter v reads 3.08v. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65A. Find the emf, the internal resistance of the battery and the resistor R. Assume that the two meters are ideal.

Here,

The voltmeter reading across the cell in open circuit is emf of cell i.e.

Emf, E = 3.08 V

Also, the voltmeter reading across the cell in closed circuit is terminal p.d. of cell i.e.

Terminal p.d, V = 3.08 V

Current, I = 1.65 A

• (i) E = ?
• (ii) Internal resistance, r = ?
• (iii) Resistance, R = ?
• (i) The voltmeter reading across the cell in open circuit is emf of cell i.e.

Emf, $E = 3.08 \text{ V}$

(ii) Using circuit formula,

$E = V + Ir$

Or, $r = \frac{E - V}{I}$

Or, $r = \frac{3.08 - 2.97}{1.65} = 0.0667 \Omega$

(iii) Using Ohm’s law,

$V = IR$

Or, $R = \frac{V}{I}$

Or, $R = \frac{2.97}{1.65}$

Or, $R = 1.8 \Omega$

Q.14. A battery of emf 1.5 V has a terminal p.d. of 1.25 V when a resistor of 25 $\Omega$ is joined to it. Calculate the current flowing, the internal resistance and terminal p.d. when a resistance and terminal p.d. when a resistance of 10 $\Omega$ replaces 25 $\Omega$ resistor.

Here,

Emf, E = 1.5 V

Terminal p.d, V = 1.25 V

$R = 25 \Omega$

When 25 $\Omega$ resistor is replaced by 25 $\Omega$ resistor,

• (i) Current, I’ = ?
• (ii) Internal resistance of cell, r = ?
• (iii) Terminal potential difference, V’ = ?

From fig. (i)

$V = IR$

$I = \frac{V}{R} = \frac{1.25}{25} = 0.05 \text{ A}$

Using circuit formula,

$E = V + Ir$

$R = \frac{E-V}{I} = \frac{1.5-1.25}{0.05} = 5 \Omega$

When 25 $\Omega$ resistor is replaced by 25 $\Omega$ resistor, Fig (ii)

Using circuit formula,

$E = V' + I'r$

$E = I'R + I'r$

$I' = \frac{E}{R+r} = \frac{1.5}{10+5} = 0.1 \text{ A}$

Again,

$V' = I' R' = 0.1 \times 10 = 1 V$

Q. 15. As shown in the figure, a battery of emf 24 V internal resistance r is connected to a circuit containing two parallel resistors of 3 $\Omega$ and 6 $\Omega$ in series with an 8 $\Omega$ resistor. The current flowing in the 3 $\Omega$ is 0.8 A. Calculate the current in the 6 $\Omega$ resistor and the internal resistance of the cell.

Ans: 0.4 A and $10 \Omega$

Here,

Emf of cell, E = 24 V

Current through $3\Omega$ , $I_1 = 0.8$ A

Current through $6\Omega$ , $I_2 = ?$

Internal resistance of cell, r = ?

Since $3\Omega$ and $6\Omega$ are in parallel combination

$\therefore$ $V_1 = V_2$

Or, $I_1 R_1 = I_2 R_2$

Or, $I_2 = \frac{I_1 R_1}{R_2} = \frac{0.8 \times 3}{6} = 0.4 \text{ A}$

$I = I_1 + I_2 = 0.8 + 0.4 = 1.2 \text{ A}$

Also, $R_{eq} = \frac{3 \times 6}{3 + 6} + 8 = 10 \Omega$

Now, using circuit formula,

$E = V + Ir$

Or, $E = I R_{eq} + Ir$

Or, $r = \frac{E - I R_{eq}}{I}$

Or, $r = \frac{24 - 1.2 \times 10}{1.2}$

Or, $r = 10 \Omega$

Heating Effect of Current

Electric power consumed (P)

$P = IV = I^2R = \frac{V^2}{R} \qquad \text{since } V = IR : I = \frac{V}{R}$

eg. Rating of an electric lamp is 50W - 220V

$P, V, R = ?$

$P = \frac{V^2}{R}$

$R = \frac{V^2}{P} \, = \frac{220^2}{50} \, = 968 \; \Omega.$

Q. 16. Two lamps rated 25W - 220V and 100W - 220V are connected to 220V supply. Calculate the powers consumed by the lamps.

Rating of 1st lamp 25W - 220V

and rating of $2^{nd}$ lamp 100W - 220V

Power consumed $P_1 = ? \& P_2 = ?$

For 1st lamp

25W-220V

$P_1 = -\frac{V_1^2}{R_1}$

$R_1 = \frac{V_1^2}{P_1} = \frac{220^2}{25} = 1936 \Omega.$

For 2^nd lamp

$P_2=\quad \frac{V_2{}^2}{R_2}$

$R_2 = \frac{V_2^2}{P_2} = \frac{220^2}{100} = 484 \Omega.$

Req. = $R_1 + R_2 = 1936 + 484 = 2420 \Omega$

$V = I R_{eq} = I = \frac{V}{R_{eq}} = \frac{220}{2420} = \frac{1}{11} A$

Now, $P_1 = I^2 R_1 = \left(\frac{1}{11}\right)^2 = 1936 = 16 \text{ W}$

& $P_2 = I^2 R_2 = \left(\frac{1}{11}\right)^2 \times 484 = 4 \text{ W}.$

Q. 17. An electric heating element to dissipate 480 watts on 240V mains is to be made from nichrome wire of 1mm diameter. Calculate the length of the wire required if the resistivity of nichrome is $1.1 \times 10^{-6}$ ohm-meter.

Given,

Power, P = 480 W

Potential, V = 240 V,

Diameter, $d = 1 \text{mm} = 1 \times 10^{-3} \text{ m}$

Length, l = ?

Resistivity, $\rho = 1.1 \times 10^{-6} \Omega m$

We have,

$R = \rho \frac{l}{A} = \rho \frac{l}{\pi d^2/4}$

Or, $l = \frac{R \pi d^2}{\rho \times 4}$ …(1)

But, $P = \frac{V^2}{R}$

Or, $R = \frac{V^2}{P} = \frac{240^2}{480} = 120 \Omega$ .

From (1)

$l = \frac{120 \times \pi (1 \times 10^{-3})^2}{1.1 \times 10^{-6} \times 4} = 85.68 \text{ m}.$

Q. 18. An electric heating element to dissipate 1.2 KW on 240V mains is to be made from Nichrome ribbon 1 mm wide and 0.05 mm thick. Calculate the length of the ribbon required if the resistivity of nichrome is $1.1 \times 10^{-6} \Omega m$ .

Power, $P = 1.2 \text{ KW} = 1.2 \times 10^3 \text{ W}$

Potential, $V = 240V$

Breath, $b = 1 \text{mm} = 1 \times 10^{-3} \text{ m}$

Thickness, $t = 0.05 \text{ mm} = 0.05 \times 10^{-3} \text{ m}$

Length, $l = ?$

Resistivity, $\rho = 1.1 \times 10^{-6} \Omega m$

We have,

$R = \rho \frac{l}{A} = \rho \frac{l}{b \times t}$

Or, $l = \frac{R \times b \times t}{\rho}$ …(1)

But,

$P = \frac{V^2}{R}$

Or, $R = \frac{V^2}{P} = \frac{240^2}{1.2 \times 10^3} = 48 \Omega$

From (1)

$l = \frac{48 \times 1 \times 10^{-3} \times 0.05 \times 10^{-3}}{1.1 \times 10^{-6}} = 2.182 \text{ m}$

Note:

If resistance of voltmeter $(V_R)$ and resistance of ammeter $(R_A)$ are neither mentioned nor asked then we consider $R_v \to \infty$ $\Omega$ and $R_A \to 0$ $\Omega$ .

But either mentioned or asked then we consider them and calculate according to laws of series and parallel combination.

Q. 19. Two resistors of resistance $1000\Omega$ and $2000\Omega$ are joined in series with a 100V supply. A voltmeter of internal resistance $4000\Omega$ is connected to measure the potential difference across $1000\Omega$ resistor. Calculate the reading shown by the voltmeter.

Ans: 28.57V

Given,

$R_1 = 1000\Omega$ , $R_2 = 2000\Omega$

Resistance of voltmeter, $R_v = 4000\Omega$

Emf, E = 100V

$R_{AB} = \frac{R_v \times R_1}{R_v + R_1} = \frac{4000 \times 1000}{4000 + 1000} = 800\Omega$

$R_{eq} = R_{AB} + R_2 = 2000 + 800 = 2800\Omega$

Using circuit formula,

or, $E = I R_{eq}$

$I = \frac{E}{R_{eq}} = \frac{100}{2800} = \frac{1}{28} A$

Now P.d. across $R_2$ , $V_2 = I R_2 = \frac{1}{28} \times 2000 = 71.43 \text{ V}$

$\therefore$ P.d. across $R_1$ or voltmeter reading

$= 100 - 71.43 = 28.57V$

Given.

$R_1 = 500\Omega$ , $R_2 = 2000\Omega$

Resistance of voltmeter, $R_v = 2000\Omega$

Emf, E = 60V

Case-1: When voltmeter is connected across $500\Omega$ resistor,

$\overline{R_{AB} = \frac{R_v \times R_1}{R_v + R_1} = \frac{2000 \times 500}{2000 + 500} = 400\Omega}$

$R_{eq} = R_{AB} + R_2 = 400 + 2000 = 2400\Omega$

Using circuit formula,

or, $E = I R_{eq}$

or, $I = \frac{E}{R_{eq}} = \frac{60}{2400} = \frac{1}{40} A$

Now P.d. across $R_2$ , $V_2 = I R_2 = \frac{1}{40} \times 2000 = 50 \text{ V}$

∴ P.d. across R₁ or voltmeter reading

$=60-50=10V$

Case-2: When voltmeter is connected across 2000Ω resistor,

$R_{BC} = \frac{R_v \times R_2}{R_v + R_2} = \frac{2000 \times 2000}{2000 + 2000} = 1000\Omega$

$R_{eq} = R_1 + R_{BC} = 500 + 1000 = 1500\Omega$

Using circuit formula,

or, $E = I R_{eq}$

$I = \frac{E}{R_{eq}} = \frac{60}{1500} = \frac{1}{25} A$

Now P.d. across $R_1$ , $V_1 = I R_1 = \frac{1}{25} \times 500 = 20 \text{ V}$

∴ P.d. across R₁ or voltmeter reading

$=60-20=40V$

Q. 21. Two resistance of $1000\Omega$ and $2000\Omega$ are placed in series with 50V mains supply. What will be the reading on a voltmeter of internal resistance 2000 $\Omega$ when placed across the 1000 $\Omega$ resistor? What fractional change in voltage occurs when voltmeter is connected? Ans: 12V, 25%

Given,

$R_1 = 1000\Omega$ , $R_2 = 2000\Omega$

Resistance of voltmeter, $R_v = 2000\Omega$

Emf, E = 50V

Case-1: When voltmeter is not connected,

$R_{eq} = R_1 + R_2 = 1000 + 2000 = 3000\Omega$

Using circuit formula,

or, $E = I R_{eq}$

$I = \frac{E}{R_{eq}} = \frac{50}{3000} = \frac{1}{60} A$

∴ P.d. across $R_1$ , $V_1 = I R_1 = \frac{1}{60} \times 1000 = \frac{50}{3} V$

Case-2: When voltmeter is connected across 1000 }\Omega\text{ resistor,} \

$$\begin{split} R_{AB} = \frac{R_v \times R_1}{R_v + R_1} = \frac{2000 \times 1000}{2000 + 1000} = \frac{2000}{3} \, \Omega \end{split}$$

$R_{eq}{}' = \ R_{AB} + R_2 \! = \! \frac{2000}{3} \, + 2000 \! = \! \frac{8000}{3} \, \Omega$

Let I’ be the current flowing in the circuit when voltmeter is connected.

Then using circuit formula,

or, $E = I' R_{eq}'$

$I' = \frac{E}{R_{eq}'} = \frac{50}{8000/3} = \frac{3}{160} A$

Now P.d. across R₂, V₂ = I’ R₂ = $\frac{3}{160}$ × 2000 = 37.5 V

$\therefore$ P.d. across $R_1 = 1000\Omega$ , $V_1' = 50 - 37.5 = 12.5V$

So fractional change in voltage = $\frac{V_1' - V_1}{V_1} = \frac{12.5 - 50/3}{50/3} = -\frac{1}{4} = \frac{1}{4}$ (in magnitude)

And % change in voltage = $\frac{V_1' - V_1}{V_1} \times 100\%$

$=\frac{1}{4} \times 100\%$ $= 25\%$

Q. 88 Twelve cells each of emf 2 V and of internal resistance 0.5 ohm are arranged in a battery of n rows and external resistance 0.4 ohm is connected to the poles of the battery. Estimate the current flowing to the resistance in terms of n.

Q.73. An electric lamp consumes 60 W at 220 V. How many dry cells of emf 1.5 V and internal resistance of $1\Omega$ are required to glow the lamp?

Here,

Power, P = 60 W

Type: 5

(Conversion of galvanometer into Ammeter & Voltmeter)

Note:

# Conversion of galvanometer into an ammeter:

Since, galvanometer and shunt are in parallel combination, ∴ p.d. across shunt (S) = p.d. across galvanometer (G)

or, $(I-Ig).S = IgG$

or, $S = \frac{Ig}{I}.G$

Fig: Conversion of galvanometer into an ammeter

# Conversion of galvanometer into voltmeter:

Maximum voltage that can be measured by voltmeter,

$V = I_g (R+G)$ or, $\frac{V}{I_g} = R+G$ or, $R = \frac{V}{I_g} - G$

Fig: Conversion of galvanometer into voltmeter

Q. 67. The resistance of the coil of galvanometer is $9.36\Omega$ and a current of 0.0224 A causes it to deflect full scale. The only shunt available has a resistance $0.025\Omega$ . What resistance must be connected in series with the coil to make it an ammeter of range 0 - 20 A?

Ans: 12.94 Ω

Given,

Resistance of galvanometer, $G = 9.36\Omega$

Current through the galvanometer for full scale deflection, $I_g = 0.0224$ A

Shunt, $S = 0.025\Omega$

$R_S = ?$

$Max^m$ current through ammeter, I = 20A

Since arm AB is parallel to arm XY

$\therefore V_{AB} = V_{XY}$

$I_g(G+R)$ $T_g(S)$

$G + R_S = \frac{(I - I_g) S}{I}$

$R_S = \frac{(I - I_g) S}{T} - G$

$=\frac{(20-0.0224)\times0.025}{0.0224}-9.36$

$= 12.94\Omega$

Q.67 & 75 are same.

Q. 81. A voltmeter coil has resistance $50 \Omega$ and a resistor of 1.15 K $\Omega$ is connected in series. It can read potential differences upto 12 volts. If the same coil is used to construct an ammeter which can measure currents upto 2A, what should be the resistance of the shunt used?

Given,

Resistance of galvanometer, $G = 50\Omega$

Series resistance, $R_S = 1.15 \text{ K} \Omega = 1.15 \times 10^3 \Omega$

Maximum voltage that can be measured by voltmeter, V = 12 V

Maximum current through ammeter, I = 2 A,

Shunt, S = ?

Case:1

From fig (a)

Maximum voltage that can be measured by voltmeter

$V = I_g (G + R_s)$

$I_g = \frac{V}{G + R_s}$

$= \frac{12}{50 + 1.15 \times 10^3}$

$I_g = 0.01 \text{ A}$

From Fig (b), Since galvanometer and Shunt are in parallel,

$\therefore$ P.d. across G = P.d. across S

$$\begin{split} & I_g G = (I - I_g) S \\ & S = \frac{I_g G}{I - I_g} = \frac{0.01 \times 50}{2 - 0.01} = 0.25\ \Omega \end{split}$$

Ans: $0.25\Omega$

A potential divider is an arrangement of resistors in series across a given p.d. to provide a known fraction of the potential difference. Fig.//// shows a potential divider with resistances $R_1$ and $R_2$ across a p.d. of V. The current flowing, I in the circuit is given by

$I = \frac{V}{R_1 + R_2}$

Then p.d. across $R_1$ , $V_1 = IR_1 = \frac{R_1}{R_1 + R_2}$

The fraction of V obtained across $R_1$ is $\frac{R_1}{R_1 + R_2}$ .

A resistor with a sliding contact can be used as a potential divider. This arrangements provides a continuously variable p.d. from zero to the full supply value.