In a potentiometer experiment when a battery of e.m.f. 2 V is included in the secondary circuit, the balanced point is 500 cm…

Q: In a potentiometer experiment when a battery of e.m.f. 2 V is included in the secondary circuit, the balanced point is 500 cm. Find the balancing of the same end when a cadmium cell of e.m.f. 1.018 V is connected to the secondary circuit.

Sol. E ∝ ℓ

$\large \frac{E_1}{E_2} = \frac{l_1}{l_2} $

$\large l_2 = \frac{E_2}{E_1} \times l_1 $

$\large l_2 = \frac{1.018}{2} \times 500 $

= 254.5 cm.

In a Potentiometer experiment the balancing length with a cell is 560 cm. When an external resistance of 10 Ω is connected…

Q: In a Potentiometer experiment the balancing length with a cell is 560 cm. When an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. Find the internal resistance of the cell.

Sol. Balancing length l1 = 560 cm

Change in balancing length (l1-l2) = 60 cm

560 – l2 = 60

∴ l2 = 500 cm

$\large r = ( \frac{l_1}{l_2} – 1 )R$

$\large r = ( \frac{560}{500} – 1 )\times 10$

r =1.2 Ω

A cell of e.m.f 2 volt and internal resistance 1.5 Ω is connected to the ends of 1 m long wire. The resistance of wire is 0.5 Ω /m….

Q: A cell of e.m.f 2 volt and internal resistance 1.5 Ω is connected to the ends of 1 m long wire. The resistance of wire is 0.5 Ω /m. Find the value of potential gradient on the wire.

Sol: $\large k = \frac{I R}{L} = ( \frac{E}{R+r} ) \frac{R}{L}$

$\large k = ( \frac{2}{0.5 + 1.5} ) \times 0.5 $

k = 0.5 V/m

Two cells A and B with same e.m.f of 2 V each and with internal resistances rA = 3.5 Ω and rB = 0.5 Ω are connected in series…

Q: Two cells A and B with same e.m.f of 2 V each and with internal resistances rA = 3.5 Ω and rB = 0.5 Ω are connected in series with an external resistance R = 3 Ω. Find the terminal voltages across the two cells.

Sol. Current through the circuit $\large i = \frac{E}{R+r}$

$\large i = \frac{2 + 2}{3 + 3.5 + 0.5} = \frac{4}{7} $

(i) R = 3 Ω , rA = 3.5 Ω , E = 2 V

Terminal voltages A , VA = E – i r

= 2 – (4/7) × 3.5 = 0 volt

(ii) rB = 0.5 Ω , R = 3 Ω , 2 V

Terminal voltage at B, VB = E-ir

= 2 – (4/7) × 0.5 = 1.714 volts.

When a battery is connected to the resistance of 10 Ω the current in the circuit is 0.12 A. The same battery give 0.07 A

Q: When a battery is connected to the resistance of 10 Ω the current in the circuit is 0.12 A. The same battery give 0.07 A current with 20 Ω. Calculate e.m.f. and internal resistance of the battery.

Sol. We know that E = Ir + IR

I1 r + I1 R1 = I2 r + I2 R2 ;

$\large r = \frac{I_2 R_2 – I_1 R_1}{I_1 – I_2} $

$\large r = \frac{0.07 \times 20 – 0.12 \times 10}{0.12 – 0.07} $

r = 0.2/0.05 = 4 Ω

Internal resistance r = 4 Ω

e.m.f. E = Ir + IR

0.12 × 4 + 0.12 × 10

= 0.48 + 1.2 ; E = 1.68 volt.

When a current drawn from a battery is 0.5 A, its terminal potential difference is 20 V. And when current drawn from it is 2.0 A…

Q: When a current drawn from a battery is 0.5 A, its terminal potential difference is 20 V. And when current drawn from it is 2.0A, the terminal voltage reduces to 16 V. Find out. E.m.f and internal resistance of the battery.

Sol. We know
V = E – Ir ; I = 0.5 A, V = 20 Volt, we have

20 = E – 0.5 r …(i)

I = 2 A , V = 16 Volt, we have

16 = E – 0.2 r …(ii)

From eqs (i) and (ii)

2 E – r = 40 and E – 2 r = 16

Solving we get E = 21.3 V , r = 2.675 Ω