Question

Class 10 physics chapter electricity

Answer 1

Question :

For the circuit shown in the diagram below :

What is the value of

• current through 6Ω resistor
• p.d. across 12Ω resistor

Solution :

❒ First of all we need to find equivalent resistance of the circuit.

A] In upper branch two resistors of 6Ω and 3Ω are connected in series so their equivalent will be 6 + 3 = 9Ω (R₁)

B] In lower branch two resistors of 12Ω and 3Ω are also connected in series so their equivalent will be 12 + 3 = 15Ω (R₂)

After solving, finally two resistors of 9Ω and 15Ω come in parallel. Final equivalent resistance of the circuit will be,

➠ 1/R = 1/9 + 1/15

➠ 1/R = (5 + 3) / 45

➠ 1/R = 8/45

➠ R = 45/8

➠ R = 5.625 Ω

❒ Net current flow in the circuit can be calculated by using ohm's formula.

⇒ V = I × R

• Voltage of battery = 4V

⇒ 4 = I × 5.625

⇒ I = 4/5.625

⇒ I = 0.7 A

We know that voltage across each branch in parallel connection is equal to the voltage of battery$.$

★ Current flow in upper branch :

➛ V = I₁ × R₁

➛ 4 = I₁ × 9

➛ I₁ = 4/9

➛ I₁ = 0.44 A

★ Current flow in lower branch :

➛ I = I₁ + I₂

➛ I₂ = I - I₁

➛ I₂ = 0.7 - 0.44

➛ I₂ = 0.26 A

Answer - 1 :

In series connection, same current flows through each resistor. Current flow in upper branch is 0.44A.

∴ Current flow through 6Ω resistor will be 0.44A.

Answer - 2 :

P.d. across 12Ω resistor is given by,

➛ p.d. = current × voltage

• Current flow in lower branch = 0.26A

➛ p.d. = 0.26 × 12

➛ p.d. = 3.12V

Answer 2

Question :

For the circuit shown in the diagram below :

What is the value of

current through 6Ω resistor

p.d. across 12Ω resistor

Solution :

❒ First of all we need to find equivalent resistance of the circuit.

A] In upper branch two resistors of 6Ω and 3Ω are connected in series so their equivalent will be 6 + 3 = 9Ω (R₁)

B] In lower branch two resistors of 12Ω and 3Ω are also connected in series so their equivalent will be 12 + 3 = 15Ω (R₂)

After solving, finally two resistors of 9Ω and 15Ω come in parallel. Final equivalent resistance of the circuit will be,

➠ 1/R = 1/9 + 1/15

➠ 1/R = (5 + 3) / 45

➠ 1/R = 8/45

➠ R = 45/8

➠ R = 5.625 Ω

❒ Net current flow in the circuit can be calculated by using ohm's formula.

⇒ V = I × R

Voltage of battery = 4V

⇒ 4 = I × 5.625

⇒ I = 4/5.625

⇒ I = 0.7 A

We know that voltage across each branch in parallel connection is equal to the voltage of battery..

★ Current flow in upper branch :

➛ V = I₁ × R₁

➛ 4 = I₁ × 9

➛ I₁ = 4/9

➛ I₁ = 0.44 A

★ Current flow in lower branch :

➛ I = I₁ + I₂

➛ I₂ = I - I₁

➛ I₂ = 0.7 - 0.44

➛ I₂ = 0.26 A

Answer - 1 :

In series connection, same current flows through each resistor. Current flow in upper branch is 0.44A.

∴ Current flow through 6Ω resistor will be 0.44A.

Answer - 2 :

P.d. across 12Ω resistor is given by,

➛ p.d. = current × voltage

Current flow in lower branch = 0.26A

➛ p.d. = 0.26 × 12

➛ p.d. = 3.12V