Circuit Physics Example 5

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Example 5

hard
A circuit has a 24 V24 \text{ V} battery with internal resistance r=1 Ωr = 1 \text{ } \Omega connected to an external resistance R=11 ΩR = 11 \text{ } \Omega. What is the current, the terminal voltage of the battery, and the power dissipated internally?

Solution

  1. 1
    Total resistance: RT=R+r=11+1=12 ΩR_T = R + r = 11 + 1 = 12 \text{ } \Omega. Current: I=ERT=2412=2 AI = \frac{\mathcal{E}}{R_T} = \frac{24}{12} = 2 \text{ A}.
  2. 2
    Terminal voltage: Vterminal=E−Ir=24−2(1)=22 VV_{\text{terminal}} = \mathcal{E} - Ir = 24 - 2(1) = 22 \text{ V}.
  3. 3
    Power lost internally: Pr=I2r=4×1=4 WP_r = I^2 r = 4 \times 1 = 4 \text{ W}. Power to external circuit: PR=I2R=4×11=44 WP_R = I^2 R = 4 \times 11 = 44 \text{ W}.

Answer

I=2 A,Vterminal=22 V,Pr=4 WI = 2 \text{ A}, \quad V_{\text{terminal}} = 22 \text{ V}, \quad P_r = 4 \text{ W}
Real batteries have internal resistance that causes the terminal voltage to be less than the EMF when current flows. Some power is wasted as heat inside the battery itself.

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