Calculate the load current (IL) in Figure 8-6 for the following values of RL: 1 kΩ, 5.6 KΩ, and 10 kΩ.

Question

Calculate the load current ([latex]I_{L}[/latex]) in Figure 8-6 for the following values of [latex] R_{L}[/latex]: 1 kΩ, 5.6 KΩ, and 10 kΩ.

Accepted Answer

For [latex]R_{L}[/latex] = 1 kΩ, the load current is

[latex]I_{L}= \left(\frac{R_{S}}{R_{S}+ R_{L}} \right)I_{S}= \left(\frac{100 \ \Omega }{101 \ \Omega } \right)1 \ A = 990 \ mA[/latex]

For [latex]R_{L}[/latex] = 5.6 kΩ,

[latex]I_{L}= \left(\frac{100 \ k \Omega }{105.6 \ k \Omega} \right) 1 \ A = 947 \ mA[/latex]

For [latex]R_{L}[/latex] = 10 kΩ,

[latex]I_{L}= \left(\frac{100 \ k \Omega }{110 \ k \Omega} \right) 1 \ A = 909 \ mA[/latex]

Notice that the load current, [latex]I_{L}[/latex], is within 10% of the source current for each value of [latex]R_{L}[/latex] because [latex]R_{L}[/latex] is at least ten times smaller than [latex]R_{S}[/latex] in each case.

Question 8.3: Calculate the load current (IL) in Figure 8-6 for the follow...

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