Answer to Question #228251 in Electric Circuits for Sakwas

Question

A battery is connected in series with a variable resistor R and an ammeter in order to form a circuit. The resistor of the ammeter may be neglected. When R is given the value 1.1ohm the current in the circuit is 4.0amp but when the R is given value 1.6ohm the current in the circuit is 3.0amp. Find the internal resistance of the battery

Answer 1

Explanations & Calculations

• Consider only a resistor diagram. Then you should visualize a circuit in which the two resistors are connected in series to the battery.
• Then we are able to apply Kirchhoff's voltage law to the circuit.
• We are not given a value for the voltage supply (neither the battery's resistance) & that is why we are given two sets of data to solve for any of them simultaneously.
• Then,

$\qquad\qquad \begin{aligned} \small V &=\small 4A\times(1.1\,\Omega+R)\cdots(1)\\\\ \small V&=\small 3A\times(1.6\,\Omega+R)\cdots(2)\\\\ \small 4(1.1+R)&=\small 3(1.6+R)\cdots\cdots(1) =\small (2)\\ \small R&=\small \bold{0.4\,\Omega} \end{aligned}$