Question

A battery having an e.m.f. of 110 V and an internal resistance of 0.2 Ω is connected in parallel with another battery having an e.m.f. of 100 V and internal resistance 0.25 Ω. The two batteries in parallelare placed in series with a regulating resistance of 5 Ω and connected across 200 V mains, as shown. Calculate :(i) The magnitude and direction of the current in each battery.(ii) The total current taken from the supply mains

Answer 1

Explanations & Calculations

• According to the description given, the assembly should be as follows

• By applying Kirchoff's voltage law in the clockwise direction as shown for both the loops

$\qquad\qquad \begin{aligned} \small 200V-110V&= \small5(i_1+i_2)\times+0.5i_1\\ \small 90&= \small 5.5i_1+ 5i_2\cdots(1)\\ \\ \small 200-100&= \small 5(i_1+i_2)+0.25i_2\\ \small 100&= \small 5i_1+5.25i_2\cdots(2) \end{aligned}$

• Solving this yeilds,

$\qquad\qquad \begin{aligned} \small i_1&=\small \bold{-7.097A\,\,:\text{opposite to the shown }}\\ \small i_2&= \small\bold{ 25.806A}\,\,\,:\text{in the shown direction} \end{aligned}$

• 110V battery is being drained while the 100V battery being charged from both the mains & the 110V one.

• The total current is what flows through the $\small 5\Omega$ resistor.

$\qquad\qquad \begin{aligned} \small i_{tot}= \small i_1+i_2&=\small -7.097+25.806\\ &=\small \bold{18.709A} \end{aligned}$