Find out the following in the electric circuit given in figure: effective resistance of two 8 Ω resistors in the combination

Since the two 8 Ω resistors are in parallel, then their effective resistance R p is given by, 1 R p = 1 R 1 + 1 R 2 1 R p = 1 8 + 1 8 ⇒ 1 R p = 2 8 ⇒ 1 R p = 1 4 Ω Hence, the effective resistance, R p = 4 Ω

From the electric circuit given in figure: Find the current flowing through 4 Ω resistor

Since two 8 Ω resistors are in parallel, then their effective resistance R p is given by 1 R p = 1 R 1 + 1 R 2 1 R p = 1 8 + 1 8 1 R p = 2 8 1 R p = 1 4 Ω R p = 4 Ω These resistors are in series with 4 Ω resistor. Total resistance in the circuit R = 4 Ω + R p = 4 Ω + 4 Ω = 8 Ω Current through the circuit I = V R = 8 8 = 1 A (Where V = Voltage of battery and R = Net equivalent resistance of circuit) Thus, current through 4 Ω resistor is 1 A as 4 Ω and R p are in series and same current flows through them.

From the electric circuit given in figure: Find the potential difference across 4 Ω resistance

Since two 8 Ω resistors are in parallel, then their effective resistance R p is given by 1 R p = 1 R 1 + 1 R 2 1 R p = 1 8 + 1 8 1 R p = 2 8 1 R p = 1 4 Ω R p = 4 Ω These resistors are in series with the 4 Ω resistor. Total resistance in the circuit R = 4 Ω + R p = 4 Ω + 4 Ω =8 Ω Current through the circuit I = V R = 8 8 = 1 A (where, V = Voltage of battery and R = Net equivalent resistance of circuit ) Thus, current through 4 Ω resistor is 1 A as 4 Ω and R p are in series and same current flows through them. Potential difference across 4 Ω resistor is a potential drop by the 4 Ω resistor i.e., V = IR = 1 × 4 = 4 V

From the electric circuit given in figure: Find the power dissipated in 4 Ω resistor.

Since two 8 Ω resistors are in parallel, then their effective resistance R p is given by 1 R p = 1 R 1 + 1 R 2 1 R p = 1 8 + 1 8 1 R p = 2 8 1 R p = 1 4 Ω R p = 4 Ω These resistors are in series with the 4 Ω resistor. Total resistance in the circuit R = 4 Ω + R p = 4 Ω + 4 Ω =8 Ω Current through the circuit I = V R = 8 8 = 1 A (where, V = Voltage of battery and R = Net equivalent resistance of circuit ) Power dissipated in 4 Ω resistor P = I 2 R = 1 2 × 4 = 4 W

From the electric circuit given in figure: Find the difference in ammeter readings, if any.

There is no difference in the readings of ammeter A 1 and A 2 as same current flows through all elements in a series circuit.

E M B I B E

NCERT EXEMPLAR PROBLEMS-SOLUTIONS SCIENCE Class X>Chapter 12 - Electricity>Short Answer Type Questions

Rajesh Singh, Indu Gupta and, Sikha Sharma Solutions for Chapter: Electricity, Exercise 2: Short Answer Type Questions

Author:Rajesh Singh, Indu Gupta & Sikha Sharma

Rajesh Singh Science Solutions for Exercise - Rajesh Singh, Indu Gupta and, Sikha Sharma Solutions for Chapter: Electricity, Exercise 2: Short Answer Type Questions

Attempt the free practice questions on Chapter 12: Electricity, Exercise 2: Short Answer Type Questions with hints and solutions to strengthen your understanding. NCERT EXEMPLAR PROBLEMS-SOLUTIONS SCIENCE Class X solutions are prepared by Experienced Embibe Experts.

Questions from Rajesh Singh, Indu Gupta and, Sikha Sharma Solutions for Chapter: Electricity, Exercise 2: Short Answer Type Questions with Hints & Solutions

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10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS SCIENCE Class X>Chapter 12 - Electricity>Short Answer Type Questions>Q 32

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?