Question

In the following circuit if ${{V}_{A}}-{{V}_{B}}=4V$ , then the value of resistance X in ohms will be:-

a)5
b)10
c)15
d)20

Answer 1

Hint: In the above question the potential difference between two points that is A and B is given to us as 60V. In the above question all the components are connected in series. Hence we will use ohm's law to determine the current in the above circuit and then again use it to determine the resistance X in the above circuit.

Formula used:
$V=IR$

Complete step-by-step answer:
Let us say there is a resistance(R) connected to a battery such that the potential difference between the two terminals is (V). Therefore the current in the circuit (I) can be found out by using the relation given to us by ohm's law i.e. $V=IR$.
Now to begin with let us first analyze the above circuit that is the upper section consisting of a resistor of 10 ohms and cell of emf 5Vand lower section consisting of resistor X and cell of emf 2V. The potential difference across the upper section is equal to the sum of potential difference across the resistor and the battery of 5V. Hence we can write,
$\begin{align}
  & {{V}_{AB}}={{V}_{(10\Omega )}}+{{V}_{(5V)}} \\
 & \Rightarrow 4V=IR+5V \\
 & \Rightarrow -1V=I10\Omega \\
 & \Rightarrow I=-\dfrac{1}{10}A \\
\end{align}$
Similarly, the potential difference across the lower section is equal to the sum of potential difference across the resistor X and the battery of 2V. All the components of the circuit are in series, hence the current in the upper part and the lower part is the same. Hence we can write,
$\begin{align}
  & {{V}_{AB}}={{V}_{(X\Omega )}}+{{V}_{(2V)}} \\
 & \Rightarrow 4V=\dfrac{1}{10}X+2V \\
 & \Rightarrow 2V=\dfrac{1}{10}X \\
 & \Rightarrow X=20\Omega \\
\end{align}$

So, the correct answer is “Option d”.

Note: It is to be noted that the value of current in the upper section of the circuit is negative while in the lower section it is positive. This is because between the same two points i.e. A and B, the current in the upper section flows from B to A while in the lower section from A to B. This can easily be understood by looking at the emf of the cells i.e. 5V > 2V hence the current will flow from the 5V cell towards the 2V cell.