Distribution of Current in Two Resistors in Parallel

Concept Explanation

Distribution of Current in Two Resistors in Parallel

In the case of a parallel combination of resistance the current is divided into different paths and the sum of current passing through each resistance is equal to the total current through the circuit. Now let us derive a relation between current and resistance to calculate the amount of current passing through each resistance. In the circuit below we have two resistance R₁ and R₂ connected in parallel.

The resistors R₁ and R₂ are connected in parallel. The current I gets distributed in the two resistors. Let I₁ and I₂ be the current flowing through the resistors R₁ and R₂ respectively. As the resistance are connected in parallel so the sum of the current flowing through two resistance is equal to the total current flowing in the circuit. So we have

$dpi{100} large I=I_{1}+I_{2}$ . (i)

Applying Ohm s law to the resistor $large R_{1}$ ,

$large V_{A}-V_{B}=R_{1}I_{1}$ (ii)

And applying Ohm s law to the resistor $large R_{2}$ ,

$large V_{A}-V_{B}=R_{2}I_{2}$ . (iii)

From (ii) and (iii) we get

$large R_{1}I_{1}=R_{2}I_{2}$

$large dpi{100} large I_{2}=frac{R_{1}}{R_{2}}I_{1}$ .

Substituting for I₂ in (i), we have

$large I=I_{1}+frac{R_{1}}{R_{2}}I_{1}=I_{1}left ( 1+frac{R_{1}}{R_{2}} right )=I_{1}frac{R_{1}+R_{2}}{R_{2}}$

or $large I_{1}=frac{R_{2}}{R_{1}+R_{2}}I$

Similarly, $large I_{2}=frac{R_{1}}{R_{1}+R_{2}}I$

Thus, $large frac{i_{1}}{i_{2}}=frac{R_{2}}{R_{1}}$

The current through each branch in a parallel combination of resistors is inversely proportional to its resistance.

Illustration: Two resistors of resistances R1 and R2 having resistance 10 ohms and 20 ohms respectively are connected in parallel. A battery supplies 6 A of current to the combination, as shown in fig. Calculate the current in each resistor.

Solution: As the two resistance are connected in parallel the current in the resistance can be calculated as follows

The current in the $large 10;Omega$ resistor is

$large I_{1}=frac{R_{2}}{R_{1}+R_{2}}I=frac{(20Omega )times(6A)}{(10Omega )+(20Omega )}=frac{120}{30} A=4 A.$

The current in the $large 20;Omega$ resistor is

$large I_{2}=frac{R_{1}}{R_{1}+R_{2}}I=frac{(10Omega )times(6A)}{(10Omega )+(20Omega )}=frac{60}{30}A=2A.$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

Two resistors are connected in series to 25 volt battery, and an ammeter in the circuit reads 0.50 A. The first resistor is rated at 22 ohms. Find :- (a) the total resistance (b) the resistance of the second resistor (c) the potential difference across the second resistor.
A. $(a);55:Omega, (b):29:Omega:(c):13 V$		B. $(a);23:Omega, (b):30:Omega:(c):10 V$
C. $(a);50:Omega, (b):28:Omega:(c):14 V$		D. $(a);60:Omega, (b):30:Omega:(c):14 V$
Right Option : C
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Question : 2

The potential difference between the terminals of an electric heater is 50 V when it draws a current of 6 A from the source.What current will the heater draw if the potential difference is increased to 110 V ?
A. 15 A	B. 17 A	C. 13.2 A	D. 5.6 A
Right Option : C
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Question : 3

In parallel combination of resistance the total current flowing through the circuit is equal to _____________________________.
A. Current passing through single resistance		B. Sum of current passing through all the resistances
C. Current cannot be calculated		D. None of the above
Right Option : B
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