Determine the voltage across the load resistor and the current through the load resistor in the bridge circuit in Figure 8-66. Notice that the resistors are labeled for convenient conversion using Equations 8-1, 8-2, and 8-3. Rc is the load resistor.

Question

Determine the voltage across the load resistor and the current through the load resistor in the bridge circuit in Figure 8-66. Notice that the resistors are labeled for convenient conversion using Equations 8-1 : [latex]R_{1}= \frac{R_{A}R_{C}}{R_{A}+ R_{B}+ R_{C}}[/latex], 8-2 : [latex]R_{2}= \frac{R_{B}R_{C}}{R_{A}+ R_{B}+ R_{C}}[/latex], and 8-3 : [latex]R_{1}= \frac{R_{A}R_{B}}{R_{A}+ R_{B}+ R_{C}}[/latex]. [latex]R_{C}[/latex] is the load resistor.

Accepted Answer

First, convert the delta formed by [latex]R_{A}[/latex],[latex]R_{b}[/latex] and [latex]R_{c}[/latex] to a wye.

[latex]R_{1}= \frac{R_{A}R_{C}}{R_{A}+ R_{B}+ R_{C}}= \frac{(2.2 \ k\Omega )(18 \ k\Omega )}{2.2 \ k\Omega +2.7 \ k\Omega + 18 \ k\Omega }= 1.73 \ k\Omega[/latex]

[latex]R_{2}= \frac{R_{B}R_{C}}{R_{A}+ R_{B}+ R_{C}}= \frac{(2.7 \ k\Omega )(18 \ k\Omega )}{22.9 \ k\Omega }= 2.12 \ k\Omega[/latex]

[latex]R_{3}= \frac{R_{A}R_{B}}{R_{A}+ R_{B}+ R_{C}}= \frac{(2.2 \ k\Omega )(2.7 \ k\Omega )}{22.9 \ k\Omega }= 259 \ \Omega [/latex]

The resulting equivalent series-parallel circuit is shown in Figure 8-67.
Next, determine [latex]R_{T}[/latex] and the branch currents in Figure 8-67.

[latex]R_{T}= \frac{(R_{1}+ R_{D})(R_{2}+ R_{E})}{(R_{1}+ R_{D})+ (R_{2}+ R_{E})} + R_{3}[/latex]

[latex]\ \ \ \ \ = \frac{(6.43 \ k\Omega )(6.02 \ k\Omega )}{6.43 \ k\Omega + 6.02 \ k\Omega} + 259 \ \Omega = 3.11 \ \Omega + 259 \ \Omega = 3.37 \ k\Omega[/latex]

[latex]I_{T}= \frac{V_{AR}}{R_{T}}= \frac{12 \ V}{3.37 \ k\Omega } = 3.56 \ mA[/latex]

The total resistance of the parallel part of the circuit, [latex]R_{T(P)}[/latex]is 3.11 [latex]k\Omega[/latex],

[latex]I_{AC}= \left(\frac{R_{T(P)}}{R_{1}+ R_{D}} \right)I_{T}= (\frac{3.11 \ k\Omega }{1.73 \ k\Omega + 4.7 \ k\Omega} )3.56 \ mA = 1.72 \ mA[/latex]

[latex]I_{AD}= \left(\frac{R_{T(P)}}{R_{2}+ R_{E}} \right)I_{T}= (\frac{3.11 \ k\Omega }{2.12 \ k\Omega + 3.9 \ k\Omega} )3.56 \ mA = 1.84 \ mA[/latex]

The voltage from terminal C to terminal D is

[latex]V_{CD}= I_{AD}R_{E}- I_{AC}R_{D}= (1.84 \ mA)(3.9 \ k\Omega )- (1.72 \ mA)(4.7 \ k\Omega )[/latex]

[latex]\ \ \ \ \ \ \ = 7.18 \ V - 8.08 \ V = - 0.9 \ V [/latex]

[latex]V_{CD}[/latex] is the voltage across the load ([latex]R_{C}[/latex]) in the bridge circuit shown in Figure 8-66. The load current through [latex]R_{C}[/latex] is

[latex]I_{R_{C}}= \frac{V_{CD}}{R_{C}}= \frac{-0.9 \ V}{18 \ k\Omega } =- 50 \ \mu A[/latex]

Question 8.21: Determine the voltage across the load resistor and the curre...

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