Question

A network resistor is connected to 16 V battery with internal resistance of 1 $\Omega$ as shown in fig.
(a)Compute the equivalent resistance of the network.
(b)Obtain the current in each resistor.
(c)Obtain the voltage drops $V_{AB},V_{BC},V_{CD}$

Answer 1

Equivalent resistance between $AB=\frac{4\times 4}{4+4}=2\Omega$

Equivalent resistance between $CD=\frac{12\times 6}{18}=4\Omega$

(a) Equivalent resistance of the network$=2+1+4+1=8\Omega$

(b) Total current in the circuit $I=\frac{V}{R}=\frac{16}{8}=2A$

• Since,the two resistors in the two arms between AB are of equal value 1A,current flows through each.
  
• The full current 2A flows the resistors BC.And one of the two resistors between CD is twice the other one.
• Therefore,current flowing through $12\Omega$ resistor is twice that of current flowing through the $6\Omega$ resistor.

Let the current flowing through the $6\Omega$ resistor be $x$,then:

$x+2x=2A$

$3x=2$

$x=\frac{2}{3}\cong 0.67A$ (through $6\Omega$ resistor)

current through $12\Omega$ resistor =$2\times \frac{2}{3}=\frac{4}{3}\cong 1.33A$

Again 2A current flows through the $1\Omega$ internal resistor of the battery.

(c) $V_{AB}=I{R}_{AB}=2\times 2=4V$

$V_{BC}=I{R}_{BC}=2\times 1=2V$

$V_{CD}=I{R}_{CD}=2\times 4=8V$