How many dry cells- each of e-m-f- 1-5 V and internal resistor 0-5- must be joined in series with a resistor of 20- to give a current of 0-6 A in the circuit-

The correct option is C 10 In series i=nEnr+R 0.6=n×1.5n×0.5+20 0.3n+12=1.5n 1.2n=12 n=121.2 n=10 ...

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Standard X

Physics

Resistors in parallel

How many dry ...

Question

How many dry cells, each of e.m.f. $1.5 V$ and internal resistor $0.5 Ω$ must be joined in series with a resistor of $20 Ω$ to give a current of $0.6 A$ in the circuit?

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Solution

The correct option is C $10$
In series
$i = \frac{n E}{n r + R}$
$0.6 = \frac{n \times 1.5}{n \times 0.5 + 20}$
$0.3 n + 12 = 1.5 n$
$1.2 n = 12$
$n = \frac{12}{1.2}$
$n = 10$

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How many dry cells, each of e.m.f. 1.5 V and internal resistor 0.5Ω must be joined in series with a resistor of 20Ω to give a current of 0.6 A in the circuit?

The correct option is C 10In seriesi=nEnr+R0.6=n×1.5n×0.5+200.3n+12=1.5n1.2n=12n=121.2n=10

How many dry cells, each of e.m.f. $1.5 V$ and internal resistor $0.5 Ω$ must be joined in series with a resistor of $20 Ω$ to give a current of $0.6 A$ in the circuit?

The correct option is C $10$
In series
$i = \frac{n E}{n r + R}$
$0.6 = \frac{n \times 1.5}{n \times 0.5 + 20}$
$0.3 n + 12 = 1.5 n$
$1.2 n = 12$
$n = \frac{12}{1.2}$
$n = 10$