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प्रश्न
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:
- the p.d. across the 4 Ω resistors,
- the p.d. across the internal resistance of the cell,
- the p.d. across the R Ω or 2 Ω resistors, and
- the value of R.
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उत्तर
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
संबंधित प्रश्न
The p.d. across a 3 Ω resistor is 6 V. The current flowing in the resistor will be:
(a) `1/2A`
(b) 1 A
(c) 2 A
(d) 6 A
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