Question

What is the value of the resistance ${{R}_{1}}$ in the following circuit?

$\text{A}\text{. }30\Omega $
$\text{B}\text{. 45}\Omega $
$\text{C}\text{. 6}0\Omega $
$\text{D}\text{. 8}0\Omega $

Answer 1

Hint: Use the Ohm’s law and find the potential difference across the resistance that is situated between the points B and C. Then find the current in resistance ${{R}_{2}}$ by the junction law. Use Ohm’s law and find the resistance ${{R}_{2}}$.
Formula used:
V = iR

Complete answer:
We can see that the current through the resistance between points B and C is 1.5A and the value of resistance is 30$\Omega $.
From Ohm’s law, we know that V = iR, where V is the potential difference across the resistance, i is the current passing in it and R is the value of the resistance.
Therefore, the potential difference across the resistance between BC is ${{V}_{2}}=(1.5)(30)=45V$.
We can see that the resistance BC in parallel connection with the resistance ${{R}_{2}}$. When two resistances are in parallel connection, the potential difference across the two is the same.
Therefore, the potential difference across ${{R}_{2}}$ is ${{V}_{2}}=45V$.
Let the current flowing through the resistance ${{R}_{2}}$ be ${{i}_{2}}$.
Now, by applying junction law at the point A we get that $2.25={{i}_{2}}+1.5$.
$\Rightarrow {{i}_{2}}=2.25-1.5=0.75A$.
From Ohm’s law we get ${{V}_{2}}={{i}_{2}}{{R}_{2}}$.
Substitute the values of ${{R}_{2}}$ and ${{V}_{2}}$.
$\Rightarrow 45=0.75{{R}_{2}}$.
$\Rightarrow {{R}_{2}}=\dfrac{45}{0.75}=60\Omega $.
It is shown that the resistance ${{R}_{1}}$ is equal to ${{R}_{2}}$.
 $\Rightarrow {{R}_{1}}={{R}_{2}}=60\Omega $.

So, the correct answer is “Option C”.

Note:
Other than just using the Ohm’s law and the junction law, we can also use the loop law. The loop law is called the Kirchhoff’s voltage law. According to the loop law, the total potential difference in any loop of the circuit is zero.