Question

A 100 ohm resistor is connected in series in a parallel combination of three parallel resistors 4 ohm 10 ohm and 20 ohm respectively and its connected to a 50 volt DC supply. Find the current in each resistor and total power dissipated.

Answer 1

As per question ,

Let the Resistors be,

$R_1=100\Omega,R_2=4\Omega,R_3=10\Omega$ and $R_4=20\Omega$

As per question,

$R_1$ is connected in series with parallel combination of $R_2,R_3,R_4$

Net resistance $R_{net}=R_1+\dfrac{1}{\dfrac{1}{R_2}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}$

$\Rightarrow R_{net}=100+\dfrac{1}{\dfrac{1}{4}+\dfrac{1}{10}+\dfrac{1}{20}}$

$\Rightarrow R_{net}=100+\dfrac{1}{\dfrac{5+2+1}{20}}$

$\Rightarrow R_{net}=100+\dfrac{20}{8}$

$\Rightarrow R_{net}=100+2.5=102.5\Omega$

Current in the circuit $i=\dfrac{V}{R_{net}}$ ( Ohm's law)

$i=\dfrac{50}{102.5}=0.4878A$

Current in resistor $R_1$ is $i_1=i=0.4878A$

Power dissipation $P_1=i_1^2R_1=(0.4878)^2(100)=23.79 watt$

Voltage drops across $R_1$ resistor $V_1=i\times R_1=0.4878\times 100=48.78V$

Now the remaining voltage $V_2=V-V_1=50-48.78=1.22V$

Since resistor $R_2,R_3,R_4$ are in parallel so Voltage remains same,

$\therefore$ Current in resistor $R_2$ is $i_2=\dfrac{1.22}{4}=0.305$ A

Power dissipation $P_2=i_2^2R_2=(0.305)^2(4)=0.3721 watt$

$\therefore$ Current in resistor $R_3$ is $i_3=\dfrac{1.22}{10}=0.122A$

Power dissipation $P_3=i_3^2R_3=(0.122)^2(10)=0.14 watt$

$\therefore$ Current in resistor $R_4$ is $i_4=\dfrac{1.22}{20}=0.061A$

Power dissipation $P_4=i_4^2R_4=(0.061)^2(20)= 0.07442 watt$