Consider a circuit in which three resistors of 10 , 20 and 30 are connected in parallel to each other to a cell of 220 V. The effective resistance and the current in the 20 resistor are _________ and _______, respectively.

The Correct Answer is: (b) $$\frac{60}{11} \, \Omega, 11 \, \text{A}$$​

Given:

Resistors: R1=10Ω,R2=20Ω,R3=30ΩR_1 = 10 \, \Omega, R_2 = 20 \, \Omega, R_3 = 30 \, \OmegaR₁=10Ω,R₂=20Ω,R₃$=30\Omega$

Voltage of the cell: $V=220\text{V}$

Effective Resistance

In a parallel circuit, the effective resistance ReffR_\text{eff}$$R_\text{eff}$$ is given by:

​$$\frac{1}{R_\text{eff}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$​​

Substitute the given values:

​$$\frac{1}{R_\text{eff}} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30}$$​​

Find the least common denominator (LCD = 60):

1Reff=660+360+260\frac{1}{R_\text{eff}} = \frac{6}{60} + \frac{3}{60} + \frac{2}{60}$$\frac{1}{R_\text{eff}} = \frac{6}{60} + \frac{3}{60} + \frac{2}{60}$$​​

Take the reciprocal to find $$R_\text{eff}$$​ :

​$$R_\text{eff} = \frac{60}{11} \,$$​​

The current $$I_2$$​through the $20\Omega$ resistor is calculated using Ohm's law:

​$$I_2 = \frac{V}{R_2}$$​​

I2=VR2I2= \frac{V}{R_2}I2=R2​V​​​Substitute the given values:

​$$I_2 = \frac{220}{20} = 11 \text{A}$$​​

Effective resistance: $$\frac{60}{11} \, \Omega$$​6011Ω\frac{60}{11} \, \Omega6011Ω\frac{60}{11} \, \Omega1160​Ω​​

Current in the $20\Omega$ resistor: $11\text{A}$

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