Determine the load currents IRL1 and IRL2 the bleeder current I3 in the two-tap loaded voltage divider in Figure 7-32.

Question

Determine the load currents [latex]I_{RL1}[/latex] and [latex]I_{RL2}[/latex] the bleeder current [latex]I_{3}[/latex] in the two-tap loaded voltage divider in Figure 7-32.

Accepted Answer

The equivalent resistance from node A to ground is the 100 kΩ load resistor [latex]R_{L1}[/latex] in parallel With the combination of [latex]R_{2}[/latex] in series with the parallel combination of [latex]R_{3}[/latex] and [latex]R_{L2}[/latex].Determine the resistance values first. [latex]R_{3}[/latex], in parallel with [latex]R_{L2}[/latex] is designated [latex]R_{B}[/latex]. The resulting equivalent circuit is shown in Figure 7-33(a).

[latex]R_{B}=\frac{R_{3}R_{L2}}{R_{3}+ R_{L2}}= \frac{(6.2 \ K\Omega ) (100 \ K\Omega )}{106.2 \ K \Omega } = 5.84 \ K \Omega [/latex]

[latex]R_{2}[/latex] in series with [latex]R_{B}[/latex] is designated[latex]R_{2+B}[/latex]. The resulting equivalent circuit is shown in Figure 7-33(b).

[latex]R_{2+B}[/latex] = [latex]R_{2}[/latex] + [latex]R_{B}[/latex]= 6.2 kΩ + 5.84 kΩ = 12.0kΩ

[latex]R_{L1}[/latex] in parallel with [latex]R_{2+B}[/latex] is designated [latex]R_{A}[/latex]. The resulting equivalent circuit is shown in Figure 7-33(c).

[latex]R_{A}=\frac{R_{L1}R_{2+B}}{R_{L1}+ R_{L2+B}}= \frac{(100 \ K\Omega ) (12.0 \ K\Omega )}{112 \ K \Omega } = 10.7 \ K \Omega [/latex]

[latex]R_{A}[/latex] is the total resistance from node A to ground. The total resistance for the circuit is

[latex]R_{T}[/latex] = [latex]R_{A}[/latex]+ [latex]R_{1}[/latex] = 10.7 kΩ + 12 kΩ = 22.7 kΩ

Determine the voltage across [latex]R_{L1}[/latex] as follows, using the equivalent circuit in Figure 7-33(c):

[latex]V_{RL1}= V_{A}= \left(\frac{R_{A}}{R_{T}} \right) V_{S}= \left(\frac{10.7 \ k\Omega }{22.7 \ k\Omega} \right)24 \ V= 11.3 \ V[/latex]

The load current through [latex]R_{L1}[/latex] is

[latex]I_{RL1}= \frac{V_{RL1}}{R_{L1}}= \left(\frac{11.3 \ V}{100 \ k\Omega } \right) = 113 \ \mu A[/latex]

Determine the voltage at node B by using the equivalent circuit in Figure 7-33(a) and the voltage at node A.

[latex]V_{B}= \left(\frac{R_{B}}{R_{2+B}} \right) V_{A}= \left(\frac{5.84 \ k\Omega }{12.0 \ k\Omega} \right)11.3 \ V= 5.50\ V[/latex]

The load current through [latex]R_{L2}[/latex] is

[latex]I_{RL2}= \frac{V_{RL2}}{R_{L2}}= \frac{V_{B}}{R_{L2}}= \frac{5.50 \ V}{100 \ K\Omega } = 55 \ \mu A[/latex]

The bleeder current is

[latex]I_{3}= \frac{V_{B}}{R_{3}}= \frac{5.50 \ V}{6.2 \ K\Omega}=887 \mu A[/latex]

Question 7.14: Determine the load currents IRL1 and IRL2 the bleeder curren...

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