Solution Electrical Numerical 15

Numerical 15: A power of 36 W is to be dissipated in a register connected across the terminals of a battery, having emf of 20V and an internal resistance of 1Ω. Find (i) What value of resistance will satisfy this condition. (ii) The terminal voltage of the battery for each of the resistances and (iii) The total power expenditure in each case.

Solution: Given:

Power dissipated (P) = 36 W

EMF (E) = 20 V

Internal Resistance (r) = 1Ω

(i) Value of resistance will satisfy this condition

Power dissipated at $\displaystyle \mathrm{R_L}$ :

$\displaystyle \mathrm{P\ =\ I_L^2 \times R_L}$ 

$\displaystyle \mathrm{I_L =\ \sqrt{\frac{36}{R_L}}}$ 

Also from figure, 

$\displaystyle \mathrm{E - I_L \left ( r + R_L \right ) = 0}$ 

$\displaystyle \mathrm{I_L = \frac{E}{r +R_L}}$ 

$\displaystyle \mathrm{I_L = \frac{20}{1 +R_L}}$ 

Thus, 

$\displaystyle \mathrm{\sqrt{\frac{36}{R_L}} = \frac{20}{1 +R_L}}$ 

solving the above equation we get: $\displaystyle \mathrm{R_L = 9 \ and\ 0.111\ ohms}$ 

(ii) The terminal voltage of the battery for each of the resistances

We know $\displaystyle \mathrm{P = \frac{V^2}{R_L}}$

or, $\displaystyle \mathrm{V = \sqrt{P\ \times R_L}}$ 

At 9Ω

$\displaystyle \mathrm{V = \sqrt{36\ \times 9}\ =\ 18\ Volts }$ 

At 0.111Ω

$\displaystyle \mathrm{V = \sqrt{36\ \times 0.111}\ =\ 1.98\ Volts }$ 

(iii) The total power expenditure in each case.

At 9Ω

$\displaystyle \mathrm{P = \frac{V^2}{(R_L + r)} }$ 

$\displaystyle \mathrm{P = \frac{20^2}{(9 + 1)} =\ 40\ W}$ 

At 0.111Ω

$\displaystyle \mathrm{P = \frac{20^2}{(0.111 + 1)} =\ 360.04\ W}$