Chapter 18, Problem 79P

(a)

To determine

Redraw the circuit to measure the current through $1.40\text{ k}\Omega$ resistor.

Answer to Problem 79P

The circuit is redrawn to measure the current through $1.40\text{ k}\Omega$ resistor and the circuit is shown below.

Explanation of Solution

The ammeters should be connected in series with the resistor to measure the current flowing thought it.

Conclusion:

The circuit is shown below,

(b)

To determine

Find the current through the $1.40\text{ k}\Omega$ resistor.

Answer to Problem 79P

The current through the $1.40\text{ k}\Omega$ resistor is $6.27\text{ mA}$ .

Explanation of Solution

The ammeter is connected in series with the $1.40\text{ k}\Omega$ resistor. Then assume that the ammeter resistance is zero. The direction of the current are shown in the below figure.

From Kirchhoff’s junction rule,

$I_1=I_2+I_3$ (I)

From Kirchhoff’s loop rule for the left mesh,

 $9.00\text{ V}-I_1\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega \right)=0$                                             (II)

From Kirchhoff’s loop rule for the right mesh,

 $I_2\left(1.40\times {10}^3\Omega \right)-I_3\left(16.0\times {10}^3\Omega \right)-I_3\left(83.0\times {10}^3\Omega \right)=0$                      (III)

Conclusion:

Solve equation III to get $I_3$

$\begin{array}{c}{I}_3=\frac{1.40\times {10}^3\Omega }{99.0\times {10}^3\Omega }{I}_2\\ =0.01414I_2\end{array}$

Substitute the above equation in equation I.

$\begin{array}{c}{I}_1=I_2+0.01414I_2\\ =1.01414I_2\end{array}$

Substitute the above equation in equation II.

$\begin{array}{c}9.00\text{ V}-1.01414I_2\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega \right)=0\\ I_2=\frac{9.00\text{ V}}{1.01414\left(35\Omega \right)+1.40\times {10}^3\Omega }\\ =6.27\text{ mA}\end{array}$

Therefore, the current through the $1.40\text{ k}\Omega$ resistor is $6.27\text{ mA}$.

(c)

To determine

Find the current through the $1.40\text{ k}\Omega$ resistor.

Answer to Problem 79P

The current through the $1.40\text{ k}\Omega$ resistor is $5.79\text{ mA}$ .

Explanation of Solution

The ammeter is connected in series with the $1.40\text{ k}\Omega$ resistor and the ammeter has the resistance of $120\Omega$. The direction of the current are shown in the below figure.

From Kirchhoff’s loop rule for the left mesh,

 $9.00\text{ V}-I_1\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega +120\Omega \right)=0$ (IV)

From Kirchhoff’s loop rule for the right mesh,

 $I_2\left(1.40\times {10}^3\Omega +120\Omega \right)-I_3\left(16.0\times {10}^3\Omega \right)-I_3\left(83.0\times {10}^3\Omega \right)=0$                   (V)

Conclusion:

Solve equation V to get $I_3$

$\begin{array}{c}{I}_3=\frac{1.40\times {10}^3\Omega +120\Omega }{99.0\times {10}^3\Omega }{I}_2\\ =0.01535I_2\end{array}$

Substitute the above equation in equation I.

$\begin{array}{c}{I}_1=I_2+0.01535I_2\\ =1.01535I_2\end{array}$

Substitute the above equation in equation IV.

$\begin{array}{c}9.00\text{ V}-1.01535I_2\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega +120\Omega \right)=0\\ I_2=\frac{9.00\text{ V}}{1.01535\left(35\Omega \right)+1.40\times {10}^3\Omega +120\Omega }\\ =5.79\text{ mA}\end{array}$

Therefore, the current through the $1.40\text{ k}\Omega$ resistor is $5.79\text{ mA}$.