Calculate the reading of each wattmeter in the circuit in Fig. 11.20 if the phase voltage at the load is 120 V and (a)ZΦ= 8 + j6Ω; (b)ZΦ= 8 -j6Ω ; (c) ZΦ= 5 + j5√3Ω; and (d)ZΦ= 10-∠75° Ω (e) Verify for (a)–(d) that the sum of the wattmeter readings equals the total power delivered to the load.

Question

Calculate the reading of each wattmeter in the circuit in Fig. 11.20 if the phase voltage at the load is 120 V and (a)[latex]Z_{\phi}= 8 + j6 \Omega[/latex]; (b)[latex]Z_{\phi}= 8 -j6 \Omega[/latex]; (c) [latex]Z_{\phi} = 5 + j5\sqrt{3}\Omega[/latex]; and (d)[latex]Z_{\phi}= 10 \angle -75^{\circ}\Omega[/latex] (e) Verify for (a)–(d) that the sum of the wattmeter readings equals the total power delivered to the load.

Accepted Answer

a)[latex]Z_{\phi}= 10 \angle 36.87^{\circ}\Omega[/latex],[latex]V_{L}=120\sqrt{3}V[/latex],and [latex]I_{L}=\frac{120}{10}=12A[/latex].

[latex]W_{1}=\left(120\sqrt{3}\right)(12)\cos\left(36.87^{\circ}+30^{\circ}\right)[/latex]

= 979.75 W,

[latex]W_{2}=\left(120\sqrt{3}\right)(12)\cos\left(36.87^{\circ}-30^{\circ}\right)[/latex]

= 2476.25 W.

b)[latex]Z_{\phi}= 10 \angle -36.87^{\circ}\Omega[/latex],[latex]V_{L}=120\sqrt{3}V[/latex],and [latex]I_{L}=\frac{120}{10}=12A[/latex].

[latex]W_{1}=\left(120\sqrt{3}\right)(12)\cos\left(-36.87^{\circ}+30^{\circ}\right)[/latex]

= 2476.25 W,

[latex]W_{2}=\left(120\sqrt{3}\right)(12)\cos\left(-36.87^{\circ}-30^{\circ}\right)[/latex]

= 979.75 W.

c)[latex]Z_{\phi}= 5\left(1+j\sqrt{3}\right)=10\angle 60^{\circ}\Omega[/latex],[latex]V_{L}=120\sqrt{3}V[/latex],and [latex]I_{L}=12A[/latex].

[latex]W_{1}=\left(120\sqrt{3}\right)(12)\cos\left(60^{\circ}+30^{\circ}\right)= 0,[/latex]

[latex]W_{2}=\left(120\sqrt{3}\right)(12)\cos\left(60^{\circ}-30^{\circ}\right)[/latex]

= 2160 W.

d)[latex]Z_{\phi}= 10 \angle -75^{\circ}\Omega[/latex],[latex]V_{L}=120\sqrt{3}V[/latex],and [latex]I_{L}=12A[/latex].

[latex]W_{1}=\left(120\sqrt{3}\right)(12)\cos\left(-75^{\circ}+30^{\circ}\right)= 1763.63 W,[/latex]

[latex]W_{2}=\left(120\sqrt{3}\right)(12)\cos\left(-75^{\circ}-30^{\circ}\right)= -645.53 W.[/latex]

e)[latex] P_{T}\left(a\right)=3(12)^{2}(8) = 3456 W, [/latex]

[latex]W_{1}+W_{2}=979.75 + 2476.25= 3456 W,[/latex]

[latex] P_{T}\left(b\right)=P_{T}\left(a\right)= 3456 W,[/latex]

[latex]W_{1}+W_{2}=2476.25 + 979.75= 3456 W,[/latex]

[latex] P_{T}\left(c\right)=3(12)^{2}(5) = 2160 W, [/latex]

[latex]W_{1}+W_{2}=0 +2160 = 2160 W,[/latex]

[latex] P_{T}\left(d\right)=3(12)^{2}(2.5882) = 1118.10 W, [/latex]

[latex]W_{1}+W_{2}= 1763.63 - 645.53 = 1118.10 W[/latex].

Question 11.6: Calculate the reading of each wattmeter in the circuit in Fi...

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