What is the voltage across [latex]C_{2}[/latex] in the circuit of Figure 12-48?

The reactance of the individual capacitors and the total reactance were determined in Example 12-18 . Substituting into Equation 12-28 : [latex]V_{\chi }= \left(\frac{V_{C\chi }}{X_{C(tot)}} \right) V_{s}[/latex]. [latex]V_{2 }= \left(\frac{V_{C2 }}{X_{C(tot)}} \right) V_{s}= \left(\frac{468 \ \Omega }{3.65 \ k\Omega } \right)10 \ V= 1.28 \ V [/latex] Notice that the voltage across the larger capacitor is the smaller fraction of the total. You can obtain the same result from Equation 12-12 : [latex]V_{\chi }= \left(\frac{C_{T }}{C_{\chi }} \right) V_{T}[/latex]. [latex]V_{2 }= \left(\frac{C_{T }}{C_{2 }} \right) V_{s}= \left(\frac{0.0087 \ \mu F }{0.068 \ \mu F } \right)10 \ V= 1.28 \ V[/latex]

Question 12.20: What is the voltage across C2 in the circuit of Figure 12-48...

Principles of Electric Circuits Electron Flow Version [EXP-40050]

What is the voltage across $C_{2}$ in the circuit of Figure 12-48?

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The reactance of the individual capacitors and the total reactance were determined in Example 12-18. Substituting into Equation 12-28 : $V_{\chi }= \left(\frac{V_{C\chi }}{X_{C(tot)}} \right) V_{s}$ .

V_{2 }= \left(\frac{V_{C2 }}{X_{C(tot)}} \right) V_{s}= \left(\frac{468 \ \Omega }{3.65 \ k\Omega } \right)10 \ V= 1.28 \ V

Notice that the voltage across the larger capacitor is the smaller fraction of the total. You can obtain the same result from Equation 12-12 : $V_{\chi }= \left(\frac{C_{T }}{C_{\chi }} \right) V_{T}$ .