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Home » The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f. 2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find: (a) the p.d. across the 4 ohm resistor (b) the p.d. across the internal resistance of the cell, (c) the p.d. across the R ohm or 2 ohm resistor, and (d) the value of R.
The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f. 2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find:
(a) the p.d. across the 4 ohm resistor
(b) the p.d. across the internal resistance of the cell,
(c) the p.d. across the R ohm or 2 ohm resistor, and
(d) the value of R.
Class 10 | Current Energy | ICSE | Physics | Selina
The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f. 2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find:
(a) the p.d. across the 4 ohm resistor
(b) the p.d. across the internal resistance of the cell,
(c) the p.d. across the R ohm or 2 ohm resistor, and
(d) the value of R.

(a) According to the question, find  p.d. across the resistor of 4 ohm.

Given,

R = 4 ohm

I = 0.25 A

Using ohm’s law

V = IR

V = 0.25 × 4

V = 1 V

(b) According to the question, find  p.d. across the internal resistance of cell.

Internal resistance r = 3 ohm

I = 0.25 A

V = IR

V = 0.25 × 3

V = 0.75 V

(c) According to the question, find  p.d. across the resistor of 2 ohm.

Effective resistance of parallel combination of 2 ohm resistances = 1 ohm

V = I / R

V = 0.25 / 1

V = 0.25 V

(d) According to the question, find  the value of R.

I = 0.25 A

ε = 2 V

r = 3 ohm

ε = I (R’ + r)

2 = 0.25 (R’ + 3)

R’ = 5 W

R = 2 ohm

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