Chapter 21: Electric Currents and Direct-Current Circuits Selected Solutions

Selected Solutions

15. Using Ohm's law, we can write

25. Recall that the electrical power is given by P = IV and gives the amount of energy (U) delivered per second. The total amount of energy delivered, then, is U = Pt = (IV)t. Therefore,

The 155-minute reserve capacity rating represents the greater amount of energy delivered by the battery.

37. Let us label the resistors as follows:

R₁ = 1.5 W, R₂ = 2.5 W, R₃ = 6.3 W, R₄ = 4.8 W, R₅ = 3.3 W, R₆ = 8.1 W

The equivalent resistance of the parallel combination of R₄, R₅, and R₆ is given by

The above parallel combination, R₄₅₆, is in series with R₃; their equivalent resistance is given by

Now, we see that R₃₄₅₆, is in parallel with R₁ and R₂. The overall equivalent resistance can be found from

45. (a) Take current, I, to flow from the positive terminal of the 12 V battery. At junction A, have I split into I₁ that flows through the 1.2 W resistor, and I₂ that flows through the 6.7 W resistor. Applying the junction rule at junction A gives

I = I₁ + I₂ ......(eq. 1)

If we apply the loop rule to the loop on the left side of the circuit in the figure (traversing it in a clockwise direction) we get

0 = 12 V - I(3.9 W) - I₁(1.2 W) - I(9.8 W).

If we solve this loop rule equation for I₁ it becomes

Applying the loop rule (clockwise) to the loop around the outside of the circuit gives

0 = 12 V - I(3.9 W) - I₂(6.7 W) - 9.0 V - I(9.8 W).

Solving for I₂ gives

Substituting eq. 2 and eq. 3 into eq. 1 gives,

From eq. 1 we can now find I₁.

From eq. 2 we can find I₂.

The currents through each resistor are as follows:

3.9 W, 0.72 A; 1.2 W: 1.8 A; 6.7 W: 1.0 A

(b) The potential at point A is greater than that at point B because the potential has been decreased by the 1.2 W resistor between A and B.

(c) By Ohm's law, the potential difference between A and B will have a magnitude of V_A - V_B = I₁(1.2 W).

47. Let's label the capacitors as follows: C₁ = 15 mF, C₂ = 8.2 mF, C₃ = 22 mF

The equivalent capacitance of the series combination of C₂ and C₃ is given by

This equivalent capacitance is in parallel with C₁. Therefore, the overall equivalent capacitance is given by

Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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