Question

A network of resistance is connected to a 16 V battery with internal resistance of 1 ohm as shown in the fig
1) Find equivalent resistance of the network
2) Find current in each resistor.
3) Find voltage drops $V_{AB},V_{BC},V_{CD}$

Answer 1

Given,

A network resistor is connected to $16V$ battery with terminal resistance $1\Omega$

i) The resistor are connected in parallel across AB,

Therefore,

$\frac{4\times 4}{4+4}=2\Omega$

Similarly,

$\frac{12\times 6}{12+6}=4\Omega$

Now,

$2\omega ,4\Omega ,1\Omega$ are connected in series,

So,

$1+2+4=7\Omega$

ii)The current in each resistor is:

$I=\frac{E}{R+r}=\frac{16}{7+1}=2A$

Now, Let us consider the resistor between C and D are the parallel combination of two resistences,

So, the current should be divided in the reverse ratio of the resistance,

if $I_1$ is the current through $12\Omega$ and $I_2$ is the current through $6\Omega$ then,

$\frac{T_1}{I_2}=\frac{6}{12}=\frac{1}{2}$

So, the current through $12\Omega$ and $6\Omega$ is $\frac{2}{3}A$and $\frac{4}{3}A$

iii) The voltage $V_{AB}$ in between A and B is the product of the total current between A and B and the equivalent resistance between A and B,

Therefore,

$V_{AB}=2\times 2=4v$

Voltage drop across $V_{BC}=2\times 1=2V$

Voltage across $V_{CD}=2\times 4=8V$