A voltmeter is connected in parallel with a variable resistance are which is in series with an ammeter and a cell as shown in figure. For one value of R, the metres read 0.3 A and 0.9 V. For another value of R, the readings are 0.25 A and 1.0 V. What is the internal resistance of the cell.
(A). $0.5\Omega $
(B). $2\Omega $
(C). $1.2\Omega $
(D). $1\Omega $
Answer
Verified
492.9k+ views
Hint: Try to understand the circuit first. Then try to make the voltage equations using the given conditions. Then by using the condition that the emf of the cell will be the same at all conditions we can find the value of the internal resistance of the cell.
Complete step by step answer:
Let, E be the emf of the cell and r be its internal resistance.
Let I is the current flowing through the circuit.
Given, R is the resistance and the voltmeter is connected in parallel to it.
Voltage equation for this circuit can be written as,
$E-Ir-IR=0\text{ }\to \text{ 1}$
Now, the voltmeter reading across the resistor is V. so, we can write,
$V=IR$
Now, equation 1 can be re-written as,
$\begin{align}
& E-Ir-V=0 \\
& \text{E=Ir+v }\to \text{ 2} \\
\end{align}$
Now, in the first case or in one value of R we have the metre readings as 0.3 A and 0.9 V.
So, equation 2 will be,
$E=0.3r+0.9\text{ }\to \text{ 3}$
In the second case we have the metre readings as 0.25 A and 0.9 V.
So,
$E=0.25r+1\text{ }\to \text{ 4}$
Since, the value of emf of the cell will be the same for both the cases we can equate the equation 3 and equation 4.
So,
$\begin{align}
\Rightarrow & 0.3r+0.9=0.25r+1 \\
\Rightarrow & 0.3r-0.25r=1-0.9 \\
\Rightarrow & 0.05r=0.1 \\
\Rightarrow & r=\dfrac{0.1}{0.05} \\
\Rightarrow & r=2\Omega \\
\end{align}$
So, the value of the internal resistance of the cell is 2 Ohms.
The correct option is (B).
Note: Internal resistance of a cell or battery can be defined as the opposition to the flow of current through the cell causing the battery to heat up. The internal resistance is measured in Ohm’s. Emf or the electromotive force can be defined as the energy per unit electric charge given by an energy source or battery.
Complete step by step answer:
Let, E be the emf of the cell and r be its internal resistance.
Let I is the current flowing through the circuit.
Given, R is the resistance and the voltmeter is connected in parallel to it.
Voltage equation for this circuit can be written as,
$E-Ir-IR=0\text{ }\to \text{ 1}$
Now, the voltmeter reading across the resistor is V. so, we can write,
$V=IR$
Now, equation 1 can be re-written as,
$\begin{align}
& E-Ir-V=0 \\
& \text{E=Ir+v }\to \text{ 2} \\
\end{align}$
Now, in the first case or in one value of R we have the metre readings as 0.3 A and 0.9 V.
So, equation 2 will be,
$E=0.3r+0.9\text{ }\to \text{ 3}$
In the second case we have the metre readings as 0.25 A and 0.9 V.
So,
$E=0.25r+1\text{ }\to \text{ 4}$
Since, the value of emf of the cell will be the same for both the cases we can equate the equation 3 and equation 4.
So,
$\begin{align}
\Rightarrow & 0.3r+0.9=0.25r+1 \\
\Rightarrow & 0.3r-0.25r=1-0.9 \\
\Rightarrow & 0.05r=0.1 \\
\Rightarrow & r=\dfrac{0.1}{0.05} \\
\Rightarrow & r=2\Omega \\
\end{align}$
So, the value of the internal resistance of the cell is 2 Ohms.
The correct option is (B).
Note: Internal resistance of a cell or battery can be defined as the opposition to the flow of current through the cell causing the battery to heat up. The internal resistance is measured in Ohm’s. Emf or the electromotive force can be defined as the energy per unit electric charge given by an energy source or battery.
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