Determine the current through each resistor and the voltage at each labeled node with respect to ground in the ladder network of Figure 7-38.

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To find the current through each resistor, you must know the total source current ( [latex]I_{T}[/latex]). To obtain [latex]I_{T}[/latex], you must find the total resistance "seen" by the source. Determine [latex]R_{T}[/latex] in a step-by-step process, starting at the right of the circuit diagram. First, notice that [latex]R_{5}[/latex] and [latex] R_{6}[/latex], are in series across [latex]R_{4}[/latex] Neglecting the circuit to the left of node B, the resistance from node B to ground is

[latex]R_{B}= \frac{R_{4}(R_{5}+ R_{6})}{R_{4}+ (R_{5}+ R_{6})}= \frac{(10 \ k\Omega )(9.4 \ k\Omega )}{19.4 \ k\Omega} = 4.85 \ k\Omega [/latex]

Using [latex]R_{B}[/latex], you can draw the equivalent circuit as shown in Figure 7-39.

Next, neglecting the circuit to the left of node A, the resistance from node A to ground ([latex]R_{A}[/latex]) is [latex]R_{2}[/latex] in parallel with the series combination of [latex]R_{3}[/latex] and [latex]R_{B}[/latex]. Calculate resistance [latex]R_{A}[/latex].

[latex]R_{A}= \frac{R_{2}(R_{3}+ R_{B})}{R_{2}+ (R_{3}+ R_{B})}= \frac{(8.2 \ k\Omega )(8.15 \ k\Omega )}{16.35 \ k\Omega} = 4.09 \ k\Omega [/latex]

Using [latex]R_{A}[/latex], you can further simplify the equivalent circuit of Figure 7-39 as shown in Figure 7-40.

Finally, the total resistance "seen" by the source is [latex]R_{1}[/latex] in series with [latex]R_{A}[/latex].

[latex]R_{T}[/latex] = [latex]R_{1}[/latex] + [latex]R_{A}[/latex] = 1.0 kΩ + 4.09 kΩ = 5.09 kΩ

The total circuit current is

[latex]I_{T}= \frac{V_{S}}{R_{T}}= \frac{45 \ V}{5.09 \ k\Omega} = 8.84 \ mA [/latex]

As indicated in Figure 7-39, [latex]I_{T}[/latex] is out of node A and consists of the currents through [latex]R_{2}[/latex] and the branch containing [latex]R_{3}[/latex] + [latex]R_{B}[/latex]. Since the branch resistances are approximately equal in this particular example, half the total current is through [latex]R_{2}[/latex] and half out of node B. So the currents through [latex]R_{2}[/latex] and [latex]R_{3}[/latex] are

[latex]I_{2}[/latex]= 4.42 mA

[latex]I_{3}[/latex]= 4.42 mA

If the branch resistances are not equal, use the current-divider formula. As indicated
in Figure 7-38.[latex]I_{3}[/latex] is out of node B and consists of the currents through [latex]R_{4}[/latex] and the branch containing [latex]R_{5}[/latex] + [latex]R_{6}[/latex]. Therefore, the currents through [latex]R_{4}[/latex]. [latex]R_{5}[/latex], and [latex]R_{6}[/latex], can be calculated.

[latex]I_{4}= \left(\frac{R_{5} + R_{6}}{R_{4}+ (R_{5} + R_{6})} \right) I_{3}= \left(\frac{9.4 \ k\Omega }{19.4 \ k\Omega} \right)4.42 \ mA= 2.14 \ mA[/latex]

[latex]I_{5}[/latex] = [latex]I_{6}[/latex] = [latex]I_{3}[/latex] - [latex]I_{4}[/latex] = 4.42 mA - 2.14 mA = 2.28 mA

To determine [latex]V_{A}[/latex], [latex]V_{H}[/latex], and [latex]V_{C}[/latex], apply Ohm's law.

[latex]V_{A} = I_{2}R_{2}= \left(4.42 \ mA\right) \left(8.2 \ k\Omega \right) = 36.2 \ V[/latex]

[latex]V_{B} = I_{4}R_{4}= \left(2.14 \ mA\right) \left(10 \ k\Omega \right) = 21.4 \ V[/latex]

[latex]V_{C} = I_{6}R_{6}= \left(2.28 \ mA\right) \left(4.7 \ k\Omega \right) = 10.7 \ V[/latex]

Question 7.16: Determine the current through each resistor and the voltage ...

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