Question

I kindly request the give the appropriate answer ​

Answer 1

We are given a circuit where R₁ , R₂ are connected in series having values 4Ω , 20Ω respectively . Voltage of the battery is 6 V

(i)

R₁ , R₂ are connected in series ,

➠ Rₑ₉ = R₁ + R₂

➠ Rₑ₉ = 4 + 20

➠ Rₑ₉ = 24 Ω  $\pink{\bigstar}$

Total resistance in the circuit is 24 Ω

(ii)

We know that , " Current across each resistor in the series connection (circuit) is same as the eq. current "

Apply Ohm's Law ,

➠ V = IRₑ₉

➠ 6 = I(24)

➠ I = ⁶/₂₄

➠ I = ¹/₄

➠ I = 0.25 Ω $\orange{\bigstar}$

This current is passing through the each and every resistor as  well as in circuit too .

➠ I₁ = I₂ = I = 0.25 Ω

(iii)

We know that , " Potential difference across the each resistor in the series connection (circuit)  varies "

We have the values of current across each resistor as well as resistance value . So , we need to apply Ohm's Law ,

Potential difference across R₁ :

➠ V₁ = I₁R₁

➠ V₁ = (0.25)(4)

➠ V₁ = 1 V  $\green{\bigstar}$

Potential difference across R₂ :

➠ V₂ = I₂R₂

➠ V₂ = (0.25)(20)

➠ V₂ = 5 V  $\purple{\bigstar}$

★══════════════════════★

Key Points :

• Current passing through each and every resistor in the series connection will be same
• I = I₁ = I₂
• Sum of voltages passing through each and every resistor in the series connection will be same as the voltage across the battery
• V = V₁ + V₂
• Voltage passing through each and every resistor in the parallel connection will be same
• V = V₁ = V₂

• Sum of currents passing through the each and every resistor in parallel connection will be equals to eq. current in the circuit
• I = I₁ + I₂

Answer 2

Answer:

We are given a circuit where R₁ , R₂ are connected in series having values 4Ω , 20Ω respectively . Voltage of the battery is 6 V

(i)

R₁ , R₂ are connected in series ,

➠ Rₑ₉ = R₁ + R₂

➠ Rₑ₉ = 4 + 20

➠ Rₑ₉ = 24 Ω \pink{\bigstar}★

Total resistance in the circuit is 24 Ω

(ii)

We know that , " Current across each resistor in the series connection (circuit) is same as the eq. current "

Apply Ohm's Law ,

➠ V = IRₑ₉

➠ 6 = I(24)

➠ I = ⁶/₂₄

➠ I = ¹/₄

➠ I = 0.25 Ω \orange{\bigstar}★

This current is passing through the each and every resistor as well as in circuit too .

➠ I₁ = I₂ = I = 0.25 Ω

(iii)

We know that , " Potential difference across the each resistor in the series connection (circuit) varies "

We have the values of current across each resistor as well as resistance value . So , we need to apply Ohm's Law ,

Potential difference across R₁ :

➠ V₁ = I₁R₁

➠ V₁ = (0.25)(4)

➠ V₁ = 1 V \green{\bigstar}★

Potential difference across R₂ :

➠ V₂ = I₂R₂

➠ V₂ = (0.25)(20)

➠ V₂ = 5 V \purple{\bigstar}★

★══════════════════════★

Key Points :

Current passing through each and every resistor in the series connection will be same

I = I₁ = I₂

Sum of voltages passing through each and every resistor in the series connection will be same as the voltage across the battery

V = V₁ + V₂

Voltage passing through each and every resistor in the parallel connection will be same

V = V₁ = V₂

Sum of currents passing through the each and every resistor in parallel connection will be equals to eq. current in the circuit

I = I₁ + I₂