Question

Two resistances \[{R_1} = 60\,\Omega \] and \[{R_2} = 70\,\Omega \] are connected with an ideal battery of emf \[\varepsilon = 120\,V\]. When a voltmeter is connected across \[{R_1}\], it reads 36 V. What will be its reading in volts, if it is connected across \[{R_2}\]?

Answer 1

Hint: Current in the circuit will not flow through the voltmeter because the resistance of the voltmeter is very high. Apply Kirchhoff’s law to express the potential across each component in the circuit. Substitute the voltage across \[{R_1}\] and emf of the battery and find out the voltage across \[{R_2}\].

Formula used:
Ohm’s law, \[V = IR\], where, V is the voltage, I is the current and R is the resistance.

Complete step by step answer:
We know that the voltmeter has high resistance. Therefore, the current in the circuit will not flow through the voltmeter. Thus, the direction of current will be from the positive terminal of the battery, \[{R_1}\], \[{R_2}\] and the negative terminal of the battery. We have given that the voltage across the resistor \[{R_1}\] is 36 V and we want to determine the voltage across the resistor C.

Applying Kirchhoff’s voltage law in the above closed circuit, we get,
\[{V_1} + {V_2} = \varepsilon \]
Here, \[{V_1}\] is the voltage across the resistor \[{R_1}\] and \[{V_2}\] is the voltage across the resistor \[{R_2}\].
Substituting 36 V for \[{V_1}\] and 120 V for \[\varepsilon \] in the above equation, we get,
\[36 + {V_2} = 120\]
\[ \Rightarrow {V_2} = 120 - 36\]
\[ \therefore {V_2} = 84\,{\text{V}}\]

Therefore, the voltage across the resistor \[{R_2}\] measured by the voltmeter is 84 V.

Note: Only in the case of ideal voltmeter, the resistance is infinite. For practical voltmeters, a small amount of current still flows through the voltmeter but we can neglect that small amount. While taking the sum of potential differences in the closed circuit, the voltage across the resistor should be taken as negative as there is a drop in the voltage across the resistor.