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Current Electricity

Question

A battery of emf 10 V and internal resistance of 0.5 ohm is connected across a variable resistance R. The maximum value of R is given by

0.5 Ω

1.00 Ω

2.0 Ω

0.25 Ω

Solution

0.5 Ω

V = E - I r

IR = E - I r

Where E is the open circuit e.m.f

I = $\frac{E}{R + r}$

= $\frac{10}{R + 0.5}$

= $\frac{20}{2 R + 1}$

P = I² R

= $\frac{{(20)}^{2} R}{{(2 R + 1)}^{2}}$

P = 400 $(\frac{{(2 R + 1)}^{2} - 4 R (2 R + 1)}{{(2 R + 1)}^{4}})$ = 0

( 2R + 1 )² = 4R

⇒ 2R = 1

⇒ R = 0.5