Question

For the given circuit, in the steady state, $\left| V _{ B ^{-}}- V _{ D }\right|=$___$V$

Hint:

In the steady state, capacitors behave like open circuits, so the voltage across the capacitor remains constant.

Answer 1

Correct Answer: 1

In steady state, capacitor behaves as an open circuit. Circuit is:

\(⇒ i_{AB}=\frac{6}{3}=2A\ \&\ i_{AD}=\frac{6}{12}=0.5A  \)
\(⇒\) V_B + 2 × 2 - 10 × 0.5 = V_D
\(⇒\) V_B + V_D = 1 volt
So, the correct answer is 1.

Answer 2

In the steady state, the capacitor in the circuit behaves like an open circuit because the voltage across the capacitor cannot change instantaneously. 
Therefore, the circuit simplifies as follows:

• The current only flows through the resistors in series and parallel.
• Apply Kirchhoff's Voltage Law (KVL) and Ohm's Law to find the voltage difference between points \( B \) and \( D \).

After solving the circuit, the voltage difference is found to be 1 V.