Question

Three resistors are connected in series across a 12 V battery. The first resistance has a value of 2ohms, second has a voltage drop of 4 V and third has a power dissipation of 12W. Calculate the value of the circuit current.

Answer 1

Given :

• There are 3 resistors connected in series across a 12V battery.
• The resistance has value of 2 ohms.
• Second has a voltage drop of 4V.
• Third has a power of dissipation of 12W.

To Calculate :

• The value of the circuit current = ?

Solution :

Let $\sf R_1, \ R_2 \ and \ R_3$ be the resistors.

Let the current in the circuit be I.

$\quad \sf V \ = \ IR$

$\quad \sf V \ = \ I (R_1 \ + \ R_2 \ + \ R_3)$

$\quad \sf V \ = \ IR_1 \ + \ IR_2 \ + \ IR_3 \ -- (i)$

Now,

• $\sf R_1 \ = \ 1 \Omega$
• $\sf IR_2 \ = \ 4 Volts.$

Here,

$\implies \sf P_3 \ = \ I^2R_3$

$\implies \sf R_3 \ = \ \dfrac {P_3}{I^2}$

Also,

• $\sf P_3 \ = \ 12 W$
• $\sf V \ = \ 12 V$

Now, Using the values of (i),

We can find the current = ?

$\implies \sf 12 \ = \ I(1) \ + \ 4 \ + \ I \bigg( \dfrac {12}{I^2} \bigg)$

$\implies \sf 8 \ = \ I \ + \ \dfrac {12}{I}$

$\implies \sf I^2 \ + \ 12 \ = \ 8 I$

$\implies \sf I^2 \ - \ 8 I \ + \ 12 \ = \ 0$

Now, solve the roots of the equation,

Now, solve the roots of the equation, We get,

$\implies \sf I \ - \ 2 \ = \ 0 \ or \ I \ - \ 6 \ = \ 0$

$\implies \sf I \ = \ 2A \ or \ I \ = \ 6A$

$\therefore$ The value of the circut current will be 2 Ampere or 6 Ampere.