Determine the voltage drop across each resistor in Figure 7-24.

Question

Accepted Answer

Because the total voltage is given in the figure, you can solve this problem using the voltage-divider formula. First, you need to reduce each parallel combination to an equivalent resistance. Since [latex] R_{1}[/latex]and [latex]R_{2}[/latex] are in parallel between A and B. combine their values.

[latex]R_{AB}= \frac{R_{1}R_{2}}{R_{1}+ R_{2}} = \frac{(3.3 \ k\Omega )(6.2 \ k\Omega )}{9.5 \ k\Omega} = 2.15 \ k\Omega[/latex]

Since [latex]R_{4}[/latex] is in parallel with the [latex]R_{5}[/latex] and [latex]R_{6}[/latex], series combination ([latex]R_{5+6}[/latex]) between C and D,combine these values.

[latex]R_{CD}= \frac{R_{4}R_{5+6}}{R_{4}+ R_{5+6}} = \frac{(1.0 \ k\Omega )(1.07 \ k\Omega )}{2.07 \ k\Omega} = 517 \ k\Omega[/latex]

The equivalent circuit is drawn in Figure 7-25. The total circuit resistance is
[latex]R_{T}[/latex] = [latex]R_{AB}[/latex] + [latex]R_{3}[/latex] + [latex]R_{CD}[/latex] = 2.15 kΩ + 1.0 kΩ + 517 Ω = 3.67 kΩ

Next, use the voltage-divider formula to determine the voltages in the equivalent circuit.

[latex]V_{AB}= \left(\frac{R_{AB}}{R_{T}} \right) V_{S} = \left(\frac{2.15 \ k\Omega}{3.67 \ k\Omega} \right) 8 \ V = 4.69 \ V[/latex]

[latex]V_{CD}= \left(\frac{R_{CD}}{R_{T}} \right) V_{S} = \left(\frac{517 \ \Omega}{3.67 \ k\Omega} \right) 8 \ V = 1.13 \ V[/latex]

[latex]V_{3}= \left(\frac{R_{3}}{R_{T}} \right) V_{S} = \left(\frac{1.0 \ k\Omega}{3.67 \ k\Omega} \right) 8 \ V = 2.18 \ V[/latex]

Refer to Figure 7-24. [latex]V_{AB}[/latex] equals the voltage across both [latex]R_{1}[/latex], and [latex]R_{2}[/latex]. so

[latex]V_{1}[/latex] = [latex]V_{2}[/latex] = [latex]V_{AB}[/latex] = 4.69 V

[latex]V_{CD}[/latex] is the voltage across [latex]R_{4}[/latex] and across the series combination of [latex]R_{5}[/latex] and [latex]R_{6}[/latex] Therefore,

[latex]V_{4}[/latex] = [latex]V_{CD}[/latex] = 1.13 V

Now apply the voltage-divider formula to the series combination of [latex]R_{5}[/latex] and [latex]R_{6}[/latex] to get [latex]V_{5}[/latex] and [latex]V_{6}[/latex]

[latex]V_{5}= \left(\frac{R_{5}}{R_{5}+ R_{6}} \right)V_{CD}= \left(\frac{680 \ \Omega }{1070 \ \Omega } \right) 1.13 \ V= 718 \ mV[/latex]

[latex]V_{6}= \left(\frac{R_{6}}{R_{5}+ R_{6}} \right)V_{CD}= \left(\frac{390 \ \Omega }{1070 \ \Omega } \right) 1.13 \ V= 412 \ mV[/latex]

Question 7.12: Determine the voltage drop across each resistor in Figure 7-...

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