Question

Two touch light batteries each having e.m.f 1.5v and an internal resistor of 1ohm are connected to a Resistance of 2ohms.calculate the current in the resistance if the cells are connected in series

Answer 1

Explanations & Calculations

• Since the batteries are connected in series, their voltages add up to ($\small 1.5V+1.5V=$ ) $\small3.0V$ & their resistances are also connected in series.
• Moreover, these two resistances are also series connected with the external resistor giving the equivalent resistance of the circuit to be ($\small 1\Omega+1\Omega+2\Omega=$) $\small 4\Omega$.
• Now the circuit can be simplified to a $\small 3.0V$ battery connected to a $\small 4\Omega$ resistor.
• Now simply using $\small V=iR$ the current flowing in the circuit can be calculated.

$\qquad\qquad \begin{aligned} \small i&=\small \frac{V}{R}=\frac{3.0V}{4\Omega}\\ &=\small\bold{0.75A=750mA} \end{aligned}$