Question

The following circuit diagram shows three resistors 2 Ω, 4 Ω and R Ω connected to a battery of e.m.f 2V and internal resistance 3 Ω. If the main current of 0.25 A flows through the circuit, find:

1. the p.d. across the 4 Ω resistors,
2. the p.d. across the internal resistance of the cell,
3. the p.d. across the R Ω or 2 Ω resistors, and
4. the value of R.

Answer 1

Given: I = 0.25 A; R = 4 Ω; r = 3 Ω

(a) p.d. across the 4 Ω

V = IR

= 0.25 × 4

= 1 Volt

(b) p.d. across r of cell

V = IR

= 0.25 × 3

= 0.75 V

(c) `1/"Req" = 1/"R" + 1/2 = 1/2 + 1/2 = 2/2 = 1`

Req = 1 Ω, I = 0.25 A

V = IR

= 1 × 0.25

= 0.25 V

(d) `1/"Req" = 1/"R" + 1/2`

`1/"Req" = (2 + "R")/"2R"`

Req = `"2R"/(2 + "R")`

Total Resistance = R + r + Req

= `4 + 3 + "2R"/("R + 2")`

`= 7 + "2R"/(2 + "R")`

R = `"V"/"I"`

∴ `7 + "2R"/(2 + "R") = 2/(1/4)`

∴ `"2R"/(2 + "R") = 2 xx 4 - 7`

∴ `"2R"/(2 + "R") = 8 - 7`

∴ 2R = (2 + R)

∴ 2R - R = 2

R = 2 Ω