Question

Three resistors 3 ohm ,4 ohm,and 12 ohm are connected in parallel. What is the effective resistance of the combination.
If the combination is connected to a battery of emf 6v and internal resistance 0.5 ohm ,find the current drawn from the battery and terminal potential difference across the battery.

Answer 1

Answer:

1/R=1/3+1/4+1/12

=4+3+1/12

=8/12

=2/3

R=3/2

=1.5ohm

I=V/R

=6/2

=3 amp

Answer 2

Given :

Three resistors connected in parallel of resistances :

=  3 , 4 and  12 ohms  

Emf ob battery :

= 6 volt

Internal resistance of battery :

= 0.5 ohm

To Find :

The current drawn from the battery and the potential difference across the battery = ?

Solution :

Let the given resistances be denoted as :

$R_{1} = 3$ ohm , $R_{2}= 4$ ohm and $R_{3}=12$ ohm

Since the formula for equivalent resistance of the resistors connected in parallel is given as :

$\frac{1}{R_{eqv}} =\frac{1}{R_{1}} +\frac{1}{R_{2}} + \frac{1}{R_{3}}$

So using this formula we get :

$\frac{1}{R_{eqv}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}$

$\frac{1}{R_{eqv}}= \frac{4+3+1}{12}$

$R_{eqv}= \frac{12}{8}$

       =1.5 ohm

Now when this battery is connected to a battery of emf 6 V and internal resistance 0.5 ohm , then the total resistance of the circuit is :

R = 1.5 + 0.5  ohm

  = 2 ohm

So the current drawn from the battery is given as :

$I = \frac {V}{R}$

So, $I = \frac{6}{2} A$

Hence $I = 3 A$

Now the terminal potential difference across the battery is :

= V - Ir

=$(6- 3\times 0.5) V$

= 4.5 V

So finally the equivalent resistance , current drawn and the terminal potential difference is 1.5 ohm , 3 a and 4.5 V respectively.