Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

Answer to Problem 100P

The value of current in battery is $25.00\text{A}$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00\text{mC}$ .

The value of capacitance is $C=5.00\mu \text{F}$ .

The emf is $\epsilon =310\text{V}$ .

The current in $50.0\Omega$ resistor is $I_{50.0\Omega }=5.00\text{A}$ .

The current in $R_2$ resistor is $I_{R_2}=5.00\text{A}$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

  $I_{\text{battery}}=I_{10.0\Omega }+I_{50.0\Omega }$ ..... (1)

Calculation:

The expression for current in $10.0\Omega$ resistor is,

  $\begin{array}{c}{I}_{10.0\Omega }=\frac{V_{10.0\Omega }}{10.0\Omega }\\ =\frac{1}{\left(10.0\Omega \right)}\left(\frac{q}{C}\right)\left(\because V_{10\Omega }=\frac{q}{C}\right)\end{array}$ ..... (2)

From equation (1) and (2),

  $I_{\text{battery}}=\frac{1}{\left(10.0\Omega \right)}\left(\frac{q}{C}\right)+I_{50.0\Omega }$

The current in battery is calculated as,

  $\begin{array}{c}{I}_{\text{battery}}=\frac{1}{\left(10.0\Omega \right)}\left(\frac{1.00\text{mC}}{5.00\mu \text{F}}\right)+\left(5.00\text{A}\right)\\ =\frac{1}{\left(10.0\Omega \right)}\left(\frac{1.00\times {10}^{-3}\text{C}}{5.00\times {10}^{-6}\text{F}}\right)+\left(5.00\text{A}\right)\\ =25.00\text{A}\end{array}$

Conclusion:

Therefore, the value of current in battery is $25.00\text{A}$ .

(b)

To determine

The value of resistances $R_1,R_2$ and $R_3$ .

Answer to Problem 100P

The value of resistances $R_1,R_2$ and $R_3$ are $0.40\Omega$ , $10.0\Omega$ and $6.67\Omega$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R_3$ , $50.0\Omega$ , $5.00\Omega$ including battery,

  $\epsilon -\left(I_{R_1}\right)R_1-\left(I_{50.0\Omega }\right)\left(50.0\Omega \right)-\left(I_{5.00\Omega }\right)\left(5.00\Omega \right)=0$ ..... (1)

Here, $I_{R_1}=I_{\text{battery}}=25.0\text{A}$

Apply loop rule to resistors $10.0\Omega$ , $50.0\Omega$ , $R_2$ including battery,

  $-\left(I_{\text{battery}}-I_{5.00\Omega }\right)\left(10.0\Omega \right)-I_{R_2}\cdot R_2+\left(I_{50.0\Omega }\right)\left(50.0\Omega \right)=0$ ..... (2)

Apply loop rule to resistors $R_1$ , $10.0\Omega$ , $R_3$ including battery,

  $\epsilon -\left(I_{R_1}\right)R_1-\left(I_{\text{battery}}-I_{5.00\Omega }\right)\left(10.0\Omega \right)-\left(I_{10.0\Omega }-I_{R_2}\right)R_3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R_1$ is calculated as,

  $\begin{array}{c}\left(310\text{V}\right)-\left(25.0\text{A}\right)R_1-\left(5.00\text{A}\right)\left(50.0\Omega \right)-\left(10.0\text{A}\right)\left(5.00\Omega \right)=0\\ \left(25.0\text{A}\right)R_1=10\text{A}\cdot \Omega \\ R_1=\frac{10}{25}\Omega \\ R_1=0.40\Omega \end{array}$

From equation (2), the resistance $R_2$ is calculated as,

  $\begin{array}{c}-\left(25.0\text{A}-5.00\text{A}\right)\left(10.0\Omega \right)-\left(5.00\text{A}\right)\cdot R_2+\left(5.00\text{A}\right)\left(50.0\Omega \right)=0\\ \left(5.00\text{A}\right)\cdot R_2=50\text{A}\cdot \Omega \\ R_2=10.0\Omega \end{array}$

From equation (3), the resistance $R_3$ is calculated as,

  $\begin{array}{c}\left(310\text{V}\right)-\left(25\text{A}\right)\left(0.40\Omega \right)-\left(25\text{A}-5\text{A}\right)\left(10.0\Omega \right)-\left(20\text{A}-5\text{A}\right)R_3=0\\ \left(100\text{A}\cdot \Omega \right)-\left(20.0\text{A}-5.00\text{A}\right)R_3=0\\ 15R_3=100\Omega \\ R_3=6.666\Omega \\ R_3\approx 6.67\Omega \end{array}$

Conclusion:

Therefore, the value of resistances $R_1,R_2$ and $R_3$ are $0.40\Omega$ , $10.0\Omega$ and $6.67\Omega$ respectively.