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) - loop method:

Step 1: Kill the capacitor branch.
Once things settle down, the capacitor is fully charged and blocks direct current. The branch with the \(4\ \mu\text{F}\) capacitor and its series \(2\ \Omega\) carries zero current, so it can be ignored while finding currents.

Step 2: One simple loop remains.
What is left is a single mesh: EMF \(10\ \text{V}\), internal resistance \(1\ \Omega\), a \(2\ \Omega\) resistor and the \(3\ \Omega\) resistor, all in series. Applying Kirchhoff's voltage law, \(10 = I(1+2+3)\), so \(I = 10/6 = 1.67\ \text{A}\) (this is the battery current).

Step 3: Terminal voltage.
Terminal voltage equals the EMF minus the drop on the internal resistance: \(V = 10 - (5/3)(1) = 8.33\ \text{V}\). (Check: it also equals the drop across the external part, \(I(2+3) = (5/3)(5) = 8.33\ \text{V}\).)

Step 4: Capacitor voltage and charge.
No current in the capacitor branch means no drop on its \(2\ \Omega\) resistor, so the capacitor sits across the full \(3\ \Omega\) voltage: \(V_C = (5/3)(3) = 5\ \text{V}\). Then \(Q = CV_C = 4\ \mu\text{F}\times 5\ \text{V} = 20\ \mu\text{C}\).
\[\boxed{I = 1.67\ \text{A},\ V = 8.33\ \text{V},\ Q = 20\ \mu\text{C}}\]

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Option 2 (cell C) - Kirchhoff's voltage law:

Step 1: What EMF means.
EMF is the total energy a cell gives to each coulomb of charge as it pushes the charge around the whole loop. On open circuit (no current), the terminal voltage reads exactly the EMF; once current flows, the internal resistance changes the terminal reading.

Step 2: Write the loop equation.
Take a clockwise current \(I\). The \(100\ \text{V}\) battery aids this direction while the \(20\ \text{V}\) cell opposes it (their positive plates face the same node). KVL gives \(100 - 20 = I(58 + 2)\), i.e. \(80 = 60\,I\), so \(I = 4/3 = 1.33\ \text{A}\).

Step 3: Terminal voltage of C by direct drop.
Because current is forced into the positive terminal of C, C is under charging. Its terminal voltage is the EMF plus the internal drop: \(V_C = 20 + I(2) = 20 + (4/3)(2) = 22.67\ \text{V}\). Equivalently, going across C's terminals through the external path \(V_C = 100 - I(58) = 100 - (4/3)(58) = 100 - 77.33 = 22.67\ \text{V}\), the same value.
\[\boxed{I = 1.33\ \text{A},\ V_C = 22.67\ \text{V}}\]