Series Battery Group with Internal Resistance

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Electric Circuits Intermediate Ohm's Law

Source: High school physics (Chinese)

Problem Sets:

ohm's law

Problem

Ten identical storage cells are connected in series. Each cell has EMF $\varepsilon_0 = 2.0 \ \text{V}$ and internal resistance $r_0 = 0.04 \ \Omega$. The series battery group is connected to an external resistor $R = 3.6 \ \Omega$.

  1. Find the current in the circuit.
  2. Find the voltage across the terminals of the battery group.
$I = 5 \ \text{A}$; $U = 18 \ \text{V}$.

Cells in series add EMFs and add internal resistances:

$$\varepsilon = 10 \varepsilon_0 = 10 \times 2.0 = 20 \ \text{V}, \qquad r = 10 r_0 = 10 \times 0.04 = 0.4 \ \Omega.$$

Current from the closed-circuit form of Ohm's law:

$$I = \dfrac{\varepsilon}{R + r} = \dfrac{20}{3.6 + 0.4} = \dfrac{20}{4.0} = 5 \ \text{A}.$$

Terminal voltage (drop across the external resistor, or EMF minus internal drop):

$$U = IR = 5 \times 3.6 = 18 \ \text{V} \quad (= \varepsilon - Ir = 20 - 5 \times 0.4).$$