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Question

Question: In the given network shown in the figure current \(1A\) and \(2A\) flows across \(6\Omega \) and\(3\...

In the given network shown in the figure current 1A1A and 2A2A flows across 6Ω6\Omega and3Ω3\Omega resistance, the internal resistance of the battery is

A. 3Ω3\Omega
B. 2Ω2\Omega
C. 4Ω4\Omega
D. 6Ω6\Omega

Explanation

Solution

The internal resistance of a cell is the obstacle faced by the electrons to flow within the two plates (anode and cathode) inside the cell. The value of the internal resistance of the cell must be very low. In an ideal case for the calculation of a complex circuit, the internal resistance of the object is considered to be zero.
As per the given data,
The current of value1A1A flows from the resistor of 6Ω6\Omega
The current of value 2A2A flows from a resistor of resistance 3Ω3\Omega which is parallel to the resistor of 6Ω6\Omega
The voltage across the circuit is 24V24V
Formula to be used,
Total current flowing I=I1+I2I={{I}_{1}}+{{I}_{2}}
The voltage drop across any closed loop V=I(R+r)V=I(R+r)

Complete answer:
The circuit given in the question can be labeled as,

The total current is some of the currents in the two branches parallel to each other
I=I1+I2 I=2+1 I=3A ...(1) \begin{aligned} & I={{I}_{1}}+{{I}_{2}} \\\ & \Rightarrow I=2+1 \\\ & \Rightarrow I=3A\ \ldots ...\left( 1 \right) \\\ \end{aligned}
From the circuit given in the question, the two resistances are connected parallel and these are connected to a resistor 2Ω2\Omega in series. So the resistance of the circuit will be
R=(R1R2R2+R2)+R3 R=(6×36+3)+2 R=2+2 R=2Ω...(2) \begin{aligned} & R=\left( \dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{2}}+{{R}_{2}}} \right)+{{R}_{3}} \\\ & \Rightarrow R=\left( \dfrac{6\times 3}{6+3} \right)+2 \\\ & \Rightarrow R=2+2 \\\ & \Rightarrow R=2\Omega \quad \ldots ...\left( 2 \right) \\\ \end{aligned}
Using Ohm’s law,
V=I(R+r)V=I\left( R+r \right)
Using the given value of voltage and value of current and total resistance of the circuit calculated above. The internal resistance of the cell can be given as,
24=3(4+r) r=84 r=4Ω \begin{aligned} & 24=3\left( 4+r \right) \\\ & \Rightarrow r=8-4 \\\ & \Rightarrow r=4\Omega \\\ \end{aligned}

Thus, the correct option showing the true value of the internal resistance used for the circuit mentioned in the question is Option C.

Note:
The internal resistance of a cell can be estimated by using an instrument named potentiometer. It is a kind of external variable resistance to a circuit. We can also control the voltage supplied to a circuit using a potentiometer and compare the internal resistance of two voltage (DC voltage) sources.