Question
At resonance, what is the relationship, including phase, for currents in the two resistors in Figure 9.4(c)?
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Consider the AC circuit containing two resistors in Figure $\mathrm{P} 22.14 .$ If the amplitude of the AC voltage source is $V_{\max }=25 \mathrm{V}, R_{1}=300 \Omega$ and $R_{2}=500 \Omega$ what is the amplitude of the current through $R_{1}$ and $R_{2} ?$
Alternating-Current Circuits and Machines
Analysis of AC Resistor Circuits
For the circuit of find $(a) I_{1} I_{2}$, and $I_{3} ;(b)$ the current in the $12-\Omega$ resistor. a) The circuit reduces at once to that shown.There we have $24 \Omega$ in parallel with $12 \Omega$, so the equivalent resistance below points- $a$ and $-b$ is $$ \frac{1}{R_{a b}}=\frac{1}{24 \Omega}+\frac{1}{12 \Omega}=\frac{3}{24 \Omega} \quad \text { or } \quad R_{a b}=8.0 \Omega $$ Adding to this the $1.0-\Omega$ internal resistance of the battery gives a total equivalent resistance of $9.0 \Omega$. To find the current from the battery, we write $$ I_{2}=\frac{\mathscr{E}}{R_{\mathrm{eq}}}=\frac{27 \mathrm{~V}}{9.0 \Omega}=3.0 \mathrm{~A} $$ This same current flows through the equivalent resistance below $a$ and $b$, and so p.d. from $a$ to $b=$ p.d. from $c$ to $d=I_{1} R_{a b}=(3.0 \mathrm{~A})(8.0 \Omega)=24 \mathrm{~V}$ Applying $V=I R$ to branch $c d$ gives Similarly, $$ \begin{array}{l} I_{2}=\frac{V_{c d}}{R_{c d}}=\frac{24 \mathrm{~V}}{24 \Omega}=1.0 \mathrm{~A} \\ I_{3}=\frac{V_{g h}}{R_{g h}}=\frac{24 \mathrm{~V}}{12 \Omega}=2.0 \mathrm{~A} \end{array} $$ As a check, note that $I_{2}+I_{3}=3.0 \mathrm{~A}=I_{1}$, as it should be. (b) Because $I_{2}=1.0 \mathrm{~A}$, the p.d. across the $2.0-\Omega$ resistor in $\underline{\text { Fig. } 28-}$ $\underline{9}(b)$ is $(1.0 \mathrm{~A})(2.0 \Omega)=2.0 \mathrm{~V}$. But this is also the p.d. across the $12-\Omega$ resistor in Fig. $28-9(a) .$ Applying $V=I R$ to the $12 \Omega$ gives $$ I_{12}=\frac{V_{12}}{R}=\frac{2.0 \mathrm{~V}}{12 \Omega}=0.17 \mathrm{~A} $$
For the circuit of Fig. $28-9(a)$, find $(a) I_{1} I_{2}$, and $I_{3} ;(b)$ the current in the $12-\Omega$ resistor. a) The circuit reduces at once to that shown in Fig. $28-9(b)$. There we have $24 \Omega$ in parallel with $12 \Omega$, so the equivalent resistance below points- $a$ and $-b$ is $$ \frac{1}{R_{a b}}=\frac{1}{24 \Omega}+\frac{1}{12 \Omega}=\frac{3}{24 \Omega} \quad \text { or } \quad R_{a b}=8.0 \Omega $$ Adding to this the $1.0-\Omega$ internal resistance of the battery gives a total equivalent resistance of $9.0 \Omega .$ To find the current from the battery, we write $$ I_{1}=\frac{\mathscr{E}}{R_{\mathrm{eq}}}=\frac{24 \mathrm{~V}}{9.0 \Omega}=3.0 \mathrm{~A} $$ This same current flows through the equivalent resistance below $a$ and $b$, and so p.d. from $a$ to $b=$ p.d. from $c$ to $d=I_{1} R_{a b}=(3.0 \mathrm{~A})(8.0 \Omega)=24 \mathrm{~V}$ Applying $V=I R$ to branch $c d$ gives $$ I_{2}=\frac{V_{c d}}{R_{c d}}=\frac{24 \mathrm{~V}}{24 \Omega}=1.0 \mathrm{~A} $$ Similarly, $$ I_{3}=\frac{V_{g h}}{R_{g h}}=\frac{24 \mathrm{~V}}{12 \Omega}=2.0 \mathrm{~A} $$ As a check, note that $I_{2}+I_{3}=3.0 \mathrm{~A}=I_{1}$, as it should be. (b) Because $I_{2}=1.0 \mathrm{~A}$, the p.d. across the $2.0-\Omega$ resistor in Fig. $28-9(b)$ is $(1.0 \mathrm{~A})(2.0 \Omega)=2.0 \mathrm{~V}$. But this is also the p.d. across the $12-\Omega$ resistor in Fig. $28-9(a)$. Applying $V=I R$ to the $12 \Omega$ gives $$ I_{12}=\frac{V_{12}}{R}=\frac{2.0 \mathrm{~V}}{12 \Omega}=0.17 \mathrm{~A} $$
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