Question

Three resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel across a battery of 12 V. Find the total current and the current through each resistor.

Answer 1

Given :

▪ Three resistors of 5Ω, 10Ω, and 30Ω are connected in parallel.

▪ Voltage of battery = 12V

To Find :

▪ Total current flow in the circuit.

▪ Current flow in each resistor.

Solution :

→ First, we have to find out equivalent resistance of the circuit after that we can calculate current flow in the circuit and in each resistor with the help of ohm's law.

→ As per ohm's law, current flow in resistor is directly proportional to the applied potential difference. (at constant temp.)

→ Formula of equivalent resistance in parallel connection is given by

☞ 1/Req = 1/R1 + 1/R2 + 1/R3

✴ Equivalent resistance :

✒ 1/Req = 1/5 + 1/10 + 1/30

✒ 1/Req = (6 + 3 + 1)/30

✒ 1/Req = 10/30

✒ Req = 30/10

✒ Req = 3Ω

✴ Net current flow :

✏ V = I × Req

✏ 12 = I × 3

✏ I = 12/3

✏ I = 4A

✴ Current flow in 5Ω resistor :

✒ V = I1 × R1

✒ 12 = I1 × 5

✒ I1 = 2.4A

✴ Current flow in 10'Ω' resistor :

✏ V = I2 × R2

✏ 12 = I2 × 10

✏ I2 = 1.2A

✴ Current flow in 30Ω resistor :

✒ V = I3 × R3

✒ 12 = I3 × 30

✒ I3 = 0.4A

Answer 2

Given :-

Three resistors are connected in a parallel

5 Ω, 10 Ω and 30 Ω

Voltage of battery (V) = 12V

To find :-

◆ The total amount of current

◆ The amount of current flow through each resistor

• Now first find the equivalent resistance

• Resistors connected in parallel

$\boxed{\sf{\dfrac{1}{R_p}= \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}....\ ..\ .. +\dfrac{1}{R_n}}}$

Where

$\sf R_p=$ Equivalent Resistance

$\longmapsto\sf \dfrac{1}{R_p}= \dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{30}\\ \\ \longmapsto\sf \dfrac{1}{R_p}= \dfrac{6+3+1}{30}\\ \\ \longmapsto\sf \dfrac{1}{R_p}= \cancel{\dfrac{10}{30}}\\ \\ \longmapsto\sf \dfrac{1}{R_p}= \dfrac{1}{3}\Omega\\ \\ \sf \ or \ Equivalent \ Resistance = 3 \Omega$

✧ Now Net current flow !

$\boxed{\sf{\star\ \ Voltage (V)= Current (I) \times Resistance (R)}}$

• We have
• Voltage (V)=12 V
• Resistance (R)= 3 ohm

◆ Put the values in the formula

$\longmapsto\sf V= IR\\ \\ \longmapsto\sf 12= I\times 3\\ \\ \longmapsto\sf \cancel{\dfrac{12}{3}}= I\\ \\ \longmapsto\sf 4= I\\ \\ \longmapsto\sf Current (I) = 4 \ Ampere$

$\bigstar{\underline{\sf{Current \ flow \ in \ 5 \Omega \ Resistor }}}$

$\longmapsto\sf V_1= I_1\times R_1\\ \\ \longmapsto\sf 12= I_1 \times 5\\ \\ \longmapsto\sf \cancel{\dfrac{12}{5}}=I_1\\ \\ \longmapsto\sf 2.4 A= I_1\\ \\ \longmapsto\sf Current_1(I_1)= 2.4\ Ampere$

$\bigstar{\underline{\sf{Current \ flow \ in \ 10 \Omega \ Resistor }}}$

$\longmapsto\sf V_2= I_2\times R_2\\ \\ \longmapsto\sf 12= I_2 \times 10\\ \\ \longmapsto\sf \cancel{\dfrac{12}{10}}=I_2\\ \\ \longmapsto\sf 1.2 A= I_2\\ \\ \longmapsto\sf Current_2(I_2)= 1.2\ Ampere$

$\bigstar{\underline{\sf{Current \ flow \ in \ 30 \Omega \ Resistor }}}$

$\longmapsto\sf V_3= I_3\times R_3\\ \\ \longmapsto\sf 12= I_3 \times 30\\ \\ \longmapsto\sf \cancel{\dfrac{12}{30}}=I_3\\ \\ \longmapsto\sf 0.4 A= I_3\\ \\ \longmapsto\sf Current_3(I_3)= 0.4\ Ampere$

$\boxed{\sf{\dag \ Total \ current \ flow = 4 \ Ampere }}$

$\boxed{\sf{\dag\ \ Current \ flow \ in \ each \ resistor= 2.4A \ \ 1.2A \ and \ 0.4A}}$