If the value of \( R \) is \( 20\ \Omega \) then the current through \( 60\ \Omega \) resistor as shown in the given figure is: (Given \( V = 10\ \mathrm{V} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0\ \mathrm{A} \)
Reading the figure: 10 Ω and 20 Ω form the upper branch, R and 40 Ω the lower branch, and the 60 Ω resistor joins the two junction points, so the network is a Wheatstone bridge with 60 Ω as the bridge arm.
The bridge is balanced when the ratios of the arms on the two sides are equal; the two junction points are then at the same potential and no current flows through the bridge arm.
\( \displaystyle \frac{10\ \Omega}{20\ \Omega} = \frac{R}{40\ \Omega} \quad\Rightarrow\quad R = 20\ \Omega \)
The given R is 20 Ω, so the bridge is balanced and the current through the 60 Ω resistor is zero, whatever the supply voltage.