The two batteries are connected in parallel, so the equivalent emf \( E_{\text{eq}} \) is calculated using the formula:
\[ E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \]
Given values: \( E_1 = 3 \, \text{V}, E_2 = 6 \, \text{V}, r_1 = 0.2 \, \Omega, r_2 = 0.4 \, \Omega \)
Substitute the values into the equation:
\[ E_{\text{eq}} = \frac{(3 \times 0.4) + (6 \times 0.2)}{0.2 + 0.4} = \frac{1.2 + 1.2}{0.6} = \frac{2.4}{0.6} = 4 \, \text{V} \]
Thus, the equivalent emf of the combination is \( 4 \, \text{V} \).
The equivalent internal resistance \( r_{\text{eq}} \) is given by:
\[ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} \]
Substitute the values:
\[ r_{\text{eq}} = \frac{(0.2 \times 0.4)}{0.2 + 0.4} = \frac{0.08}{0.6} = 0.1333 \, \Omega \]
Thus, the equivalent internal resistance of the combination is \( 0.1333 \, \Omega \).
The total resistance in the circuit is the sum of the internal resistance \( r_{\text{eq}} \) and the external resistor \( R = 4 \, \Omega \). Thus, the total resistance \( R_{\text{total}} \) is:
\[ R_{\text{total}} = r_{\text{eq}} + R = 0.1333 + 4 = 4.1333 \, \Omega \]
Using Ohm's law, the current \( I \) drawn from the combination is:
\[ I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{4}{4.1333} = 0.968 \, \text{A} \]
Thus, the current drawn from the combination is \( 0.968 \, \text{A} \).
In the given circuit, the potential difference across the 5 \(\mu\)F capacitor is
In the following diagram, the work done in moving a point charge from point P to point A, B and C are \( W_A, W_B, W_C \) respectively. Then (A, B, C are points on semicircle and point charge \( q \) is at the centre of semicircle)