Question

Find the current supplied by the battery in the network shown in the figure. Also calculate the terminal voltage across the cell and the charge on the capacitor.
OR
What do you mean by electromotive force (EMF) of a cell? In the circuit shown, calculate the terminal voltage of the cell C. The battery B is of negligible internal resistance.

Hint:

Main network: a steady-state capacitor passes no current, leaving a simple series loop; capacitor voltage equals the drop across the 3 Ω resistor. OR part: the two cells oppose, so net EMF = 100 − 20 V and cell C is being charged, giving \( V_C = E + Ir \).

Answer 1

Option 1 (main network):

Step 1: Behaviour of the capacitor in steady state.
In the steady state a fully charged capacitor allows no current to pass. So the branch containing the 4 µF capacitor and the lower 2 Ω resistor carries no current.

Step 2: Reduce the circuit.
The current therefore flows only in the loop made of the 10 V cell (internal resistance 1 Ω) in series with the 2 Ω resistor, joined across the 3 Ω resistor. All these are effectively in series in this single loop.
\[ R_{total} = 1 + 2 + 3 = 6\ \Omega \]

Step 3: Current supplied by the battery.
\[ I = \frac{E}{R_{total}} = \frac{10}{6} = 1.67\ \text{A} \;\left(=\tfrac{5}{3}\ \text{A}\right) \]

Step 4: Terminal voltage of the cell.
\[ V = E - I r = 10 - \frac{5}{3}\times 1 = 10 - 1.67 = 8.33\ \text{V} \;\left(=\tfrac{25}{3}\ \text{V}\right) \]

Step 5: Voltage across the capacitor.
The capacitor branch is connected across the 3 Ω resistor. Since no current flows in that branch, there is no potential drop across the lower 2 Ω resistor, so the full voltage across the 3 Ω resistor appears across the capacitor:
\[ V_C = I\times 3 = \frac{5}{3}\times 3 = 5\ \text{V} \]

Step 6: Charge on the capacitor.
\[ Q = C\,V_C = 4\ \mu\text{F}\times 5\ \text{V} = 20\ \mu\text{C} \]
\[\boxed{I = 1.67\ \text{A},\quad V = 8.33\ \text{V},\quad Q = 20\ \mu\text{C}}\]

---

Option 2 (cell C):

Step 1: Meaning of EMF.
The electromotive force (EMF) of a cell is the work done by the cell in driving a unit positive charge round the complete circuit, i.e. the energy supplied by the cell per coulomb of charge. It equals the potential difference across the cell's terminals when no current is drawn (open circuit).

Step 2: Identify the EMFs and resistances.
Cell C: \( E_C = 20\ \text{V} \), internal resistance \( r = 2\ \Omega \). Battery B: \( E_B = 100\ \text{V} \), negligible internal resistance. External resistor \( = 58\ \Omega \). In the figure the positive terminals of both C and B face the same (left) junction, so the two cells oppose each other.

Step 3: Net EMF and current.
\[ E_{net} = E_B - E_C = 100 - 20 = 80\ \text{V} \]
\[ R_{total} = 58 + 2 = 60\ \Omega \]
\[ I = \frac{E_{net}}{R_{total}} = \frac{80}{60} = \frac{4}{3} = 1.33\ \text{A} \]

Step 4: Terminal voltage of cell C.
Since the stronger 100 V battery drives the current backwards through cell C, cell C is being charged; current enters its positive terminal. For a cell that is being charged the terminal voltage is greater than the EMF:
\[ V_C = E_C + I r = 20 + \frac{4}{3}\times 2 = 20 + 2.67 = 22.67\ \text{V} \;\left(=\tfrac{68}{3}\ \text{V}\right) \]
\[\boxed{I = 1.33\ \text{A},\quad V_C = 22.67\ \text{V}}\]