A hostel keeps a lamp on a small battery pack…

A hostel keeps a lamp on a small battery pack. When one old cell failed, the warden replaced it with a fresher cell of higher emf and left the other in place. A student joins the two cells in parallel across the lamp, as the figure shows. Cell 1 has emf $20.0\ \text{V}$ and internal resistance $1.0\ \Omega$ . Cell 2 has emf $8.0\ \text{V}$ and internal resistance $2.0\ \Omega$ . Both positive terminals are joined to the top wire. The lamp has a steady resistance of $3.0\ \Omega$ , and the connecting wires have no resistance.

A circuit with three branches joined between a top wire and a bottom wire. The left branch holds cell 1 of emf 20 V, with its positive terminal up, in series with a resistor marked r1 = 1.0 ohm for its internal resistance. The middle branch holds cell 2 of emf 8.0 V, also with its positive terminal up, in series with a resistor marked r2 = 2.0 ohm. The right branch is the lamp, a resistor of 3.0 ohm. No currents are marked.

She expects both cells to push current into the lamp, but she wants to test this with Kirchhoff's rules. Write the current in each cell as positive when it leaves the positive terminal of that cell, so a negative value tells you the cell is being driven backwards. Give each current to two decimal places, and show how you reach the potential difference across the lamp before you find any current.

What is the potential difference across the lamp?
What is the current in cell 1?
What is the current in cell 2? Give its sign.
What power is dissipated in the lamp?
What total power is dissipated in the two internal resistances?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Let $V$ be the potential difference between the top and bottom wires. It is the same across every branch. For a cell whose current $I$ leaves its positive terminal, $V = E - Ir$ , so $I = \dfrac{E - V}{r}$ .

(a) The lamp. The junction rule gives $I_1 + I_2 = \dfrac{V}{R}$ , that is $\dfrac{20.0 - V}{1.0} + \dfrac{8.0 - V}{2.0} = \dfrac{V}{3.0}$ . Collecting the terms in $V$ , $V\left(\dfrac{1}{1.0} + \dfrac{1}{2.0} + \dfrac{1}{3.0}\right) = \dfrac{20.0}{1.0} + \dfrac{8.0}{2.0}$ . This gives $V = 13.09\ \text{V}$ .

(b) Cell 1. $I_1 = \dfrac{20.0 - 13.09}{1.0}\ \text{A} = 6.91\ \text{A}$ .

(c) Cell 2. $I_2 = \dfrac{8.0 - 13.09}{2.0}\ \text{A} = -2.55\ \text{A}$ . The sign is negative. The potential difference across cell 2 is $13.09\ \text{V}$ , which is more than its emf of $8.0\ \text{V}$ , so current is forced into its positive terminal and cell 2 is being charged by cell 1.

(d) The lamp. $P = \dfrac{V^2}{R} = \dfrac{(13.09\ \text{V})^2}{3.0\ \Omega} = 57.1\ \text{W}$ .

(e) Internal resistances. $I_1^2 r_1 + I_2^2 r_2 = (6.91)^2 \times 1.0 + (-2.55)^2 \times 2.0 = 60.7\ \text{W}$ .

Answer: (a) $13.09\ \text{V}$ . (b) $6.91\ \text{A}$ . (c) $-2.55\ \text{A}$ . (d) $57.1\ \text{W}$ . (e) $60.7\ \text{W}$ .

Check. Solve the two loop equations and the junction equation together and you get the same three currents. Energy also balances. The two emfs supply $E_1 I_1 + E_2 I_2 = 117.8\ \text{W}$ , and the lamp and the internal resistances take $57.1\ \text{W} + 60.7\ \text{W}$ , the same amount.

Watch out. Cell 2 does not push current into the lamp. It takes $20.4\ \text{W}$ of electrical power from cell 1 and stores it. If you assume both cells discharge, you get a wrong value of $V$ .

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