1.

Resistors \( R_{1}, R_{2} \) and \( R_{3} \) having values \( 5 \Omega, 10 \Omega \), and \( 30 \Omega \) respectively are connected in parallel across a battery of 12 volt Calculate(a) the current through each resistor(b) the total current in the circuit and(c) the total circuit resistance

Answer»
According to the question,
 
Resistors R1,R2 and R3 having values 5Ω, 10Ω and 30Ω are connected in parallel across a battery of 12 V. 

(a) We know that when resistors are connected in parallel combination, the total current flowing through the circuit is divided amongst the separate branches of the parallel combination, while the potential difference remains the same for the whole circuit. 


Potential difference = 12 V

Let current through R1  be I1 . Then, 
I1 = V/R1 = 12/5 = 2.4 A 

Let current through R2 be I2 . Then, 
I2 = V/R2 = 12/10 = 1.2 A 

Let current through R3 be I3 . Then, 
I3 = V/R3 = 12/30 = 0.4 A 

(b) We know that the total current flowing through a parallel combination of resistors is equal to the sum of the separate currents through each branch of the parallel combination. 

Let the total current flowing through the circuit be I. Then, 
I = I1 + I2 + I3 
I = 2.4 + 1.2 + 0.4
I = 4 A 

Therefore, the total current flowing though the parallel combination of resistors is 4 A. 

(c) We know that the reciprocal of the total circuit resistance of a parallel combination of resistors is equal to the sum of the reciprocals of the individual resistances in the parallel combination. 

Let the total circuit resistance of the parallel combination be Rp. Then, 

1/Rp = 1/R1 + 1/R2 + 1/R3 

1/Rp = 1/5 + 1/10 + 1/30

1/Rp = (6+3+1) / 30

1/Rp = 10/30 

1/Rp = 1/3 

Rp = 3Ω

Therefore, the total circuit resistance of the parallel combination of resistors is 3Ω. 


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