Question
In the circuit in Fig. $29-5, I_{1}=0.20 \mathrm{~A}$ and $R=5.0 \Omega .$ Find $\mathscr{E}$.We write the loop equation for loop cdefc:$$-R I_{1}+12.0-8.0-7.0 I_{2}=0 \quad \text { or } \quad-(5.0)(0.20)+12.0-8.0-7.0 I_{2}=0$$from which $I_{2}=0.43$ A. We can now find $I_{3}$ by applying the node rule at $e$ :$$I_{1}+I_{3}=I_{2} \quad \text { or } \quad I_{3}=I_{2}-I_{1}=0.23 \mathrm{~A}$$Now apply the loop rule to loop cdeac:$$-(5.0)(0.20)+12.0+(2.0)(0.23)+\mathscr{E}=0$$from which $\mathscr{E}=-11.5 \mathrm{~V}$. The minus sign tells us that the polarity of the battery is actually the reverse of that shown.
Step 1
The loop rule states that the sum of the potential differences (voltages) around any closed loop or mesh in a network is always equal to zero. This gives us the equation: \[ -R I_{1}+12.0-8.0-7.0 I_{2}=0 \quad \text { or } \quad-(5.0)(0.20)+12.0-8.0-7.0 Show more…
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In the circuit in Fig. 29-5, $I_{1}=0.20 \mathrm{~A}$ and $R=5.0 \Omega$. Find $\varepsilon$. We write the loop equation for loop cdefc: $$ -R I_{1}+12.0-8.0-7.0 I_{2}=0 \quad \text { or } \quad-(5.0)(0.20)+12.0-8.0-7.0 I_{2}=0 $$ from which $I_{2}=0.43$ A. We can now find $I_{3}$ by applying the node rule at $e$ : $$ I_{1}+I_{3}=I_{2} \quad \text { or } \quad I_{3}=I_{2}-I_{1}=0.23 \mathrm{~A} $$ Now apply the loop rule to loop cdeac: $$ -(5.0)(0.20)+12.0+(2.0)(0.23)+\varepsilon=0 $$ from which $\varepsilon=-11.5 \mathrm{~V}$. The minus sign tells us that the polarity of the battery is actually the reverse of that shown.
Assume the resistance values are $R_{1}=2400 \Omega, R_{2}=1400 \Omega, R_{3}=4500 \Omega,$ and $R_{4}=6000 \Omega,$ and the battery emfs are $\varepsilon_{1}=1.5 \mathrm{V}$ and $\varepsilon_{2}=3.0 \mathrm{V},$ unless stated otherwise. Use Kirchhoff's rules to analyze the circuit in Figure $\mathrm{P} 19.53$ a. Let $I_{1}$ be the branch current through $R_{1}$ and $I_{2}$ be the branch current through $R_{2}$. Write Kirchhoff's loop rule relation for a loop that travels through battery $1,$ resistor 1 , and battery 2 b. Write Kirchhoff's loop rule relation for a loop that travels through battery 2 and resistor 2 c. You should now have two equations and two unknowns $\left(I_{1} \text { and } I_{2}\right)$ Solve for the two branch currents.
Electric Currents and Circuits
DC Circuits: Batteries, Resistors, and Kirchhoff’s Rules
For the circuit shown in Fig. 29-2, find $I_{1}, I_{2}$, and $I_{3}$ if switch $S$ is $(a)$ open and $(b)$ closed. (a) When $S$ is open, $I_{3}=0$, because no current can flow through the middle branch. Applying the node rule to point $-a$ $$ I_{1}+I_{3}=I_{2} \quad \text { or } \quad I_{2}=I_{1}+0=I_{1} $$ Applying the loop rule to the outer loop acbda yields $$ -12.0+7.0 I_{1}+8.0 I_{2}+9.0=0 $$ To understand the use of signs, remember that current always flows from high to low potential through a resistor. Because $I_{2}=I_{1}$, Eq. (1) becomes $$ \begin{array}{lll} 15.0 I_{1}=3.0 & \text { or } & I_{1}=0.20 \mathrm{~A} \end{array} $$ Also, $I_{2}=I_{1}=0.20 \mathrm{~A}$. Notice that this is the same result that one would obtain by replacing the two batteries by a single $3.0$ - $\mathrm{V}$ battery. (b) With $S$ closed, $I_{3}$ is no longer necessarily zero. Applying the node rule to point- $a$ gives $$ I_{1}+I_{3}=I_{2} $$ Applying the loop rule to loop acba $$ -12.0+7.01 I_{1}-4.0 I_{3}=0 $$ and to loop adba gives $$ -9.0-8.0 I_{2}-4.0 I_{3}=0 $$ Applying the loop rule to the remaining loop, acbda, would yield a redundant equation, because it would contain no new voltage change. Now solve Eqs. (2), (3), and (4) for $I_{1}, I_{2}$, and $I_{3}$. From Eq. (4), $$ I_{3}=-2.0 I_{2}-2.25 $$ Substituting this in Eq. (3) yields $$ -12.0+7.0 I_{1}+9.0+8.0 I_{2}=0 \quad \text { or } \quad 7.0 I_{1}+8.0 I_{2}=3.0 $$ Substituting for $I_{3}$ in Eq. (2) also gives $$ \begin{array}{lll} I_{1}-2.0 I_{2}-2.25-I_{2} & \text { or } & I_{1}=3.0 I_{2}+2.25 \end{array} $$ Substituting this value in the previous equation finally leads to $$ 21.0 I_{2}+15.75+8.0 I_{2}=3.0 \quad \text { or } \quad I_{2}=-0.44 \mathrm{~A} $$ Using this in the equation for $I_{1}$, $$ I_{1}=3.0(-0.44)+2.25=-1.32+2.25=0.93 \mathrm{~A} $$ Notice that the minus sign is a part of the value we have found for $I_{2}$. It must be carried along with its numerical value. Now use (2) to find $$ I_{3}=I_{2}-I_{1}=(-0.44)-0.93=-1.37 \mathrm{~A} $$
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