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.