Question

A simple electrical circuit comprises a battery and two 10.0 ? resistors connected in series. The battery generates a voltage of 8.95 V and the current passing through the circuit is 425 mA.
(a) Calculate the internal resistance (r) of the battery.
(b) Briefly explain why a good voltmeter should have a high resistance.

Answer 1

a) Here, it is given that in a circuit we have two identical resistors in series and voltage across the circuit is given and current also. We have to find the internal resistance of the battery.

In a circuit, if a battery is connected with resistors in serious then that internal resistance of battery will also in series.

Let say internal resistance of batttery is "r"

As per ohms law

V = I * Re

Re= Effective resistnace

V = I * (R1 + R2 + r)....................(1)

As we know V = 8.95, I = 425mA or 0.425 A and R1 and R2 = 10 ohms putting in eq (1)

8.95 = 0.425 * (10+10+r)

r = (8.95/0.425)-20

r = 21.0 - 20

r = 1 ohms

(b) As we know voltmeter is always connected parallel in all types of circuit. Voltmeter whose working is to calculate potential in the circuit also has its own internal circuit. For better results and reading or better accuracy, we want that less current to be wasted in voltmeter i.e less current should be pass through the voltmeter. Now, as per ohms law current is inversely proportional to resistance, so, we have to increase the internal resistance of voltmeter so that current can be obstructed. This may lead to better accuracy in results.