Answer

1. Find the equivalent resistance = R_e = ?

   For parallel combination:

   Given Data:

   • R₁ = 6 Ω; R₂ = 10 Ω; R₃ = 18 Ω

   Formula:

   1/R_e = 1/R₁ + 1/R₂ + 1/R₃ ⇒ 1/R_e = 1/6 + 1/10 + 1/18

   ⇒ 1/R_e = (15 + 9 + 5)/90 ⇒ 1/R_e = 29/90 ⇒ R_e = 90/29 = 3.1 Ω

   So, the equivalent resistance = 3.1 Ω.

2. Calculate the current passing through each device

   For parallel combination

   Given Data:

   V = 12 V; R₁ = 6 Ω; R₂ = 10 Ω; R₃ = 18 Ω; R_e = ?

   By the ohm’s law: V = IR

   Current passing through each device:

   I₁ = V/R₁ = 12/6 = 2 A

   I₂ = V/R₂ = 12/10 = 1.2 A

   I₃ = V/R₃ = 12/18 = 0.67 A

Ans. (a) 3.1 Ω, (b) 2 A, 1.2 A, 0.67 A

Answer

The resistors are R₁ = 8 Ω, R₂ = 12 Ω and R₃ = 18 Ω.

The formula for the total resistance R_total of resistors in parallel is:

1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Plugging in the values:

1/R_total = 1/8 + 1/12 + 1/18 ⇒ 1/R_total = (9 + 6 + 4)/72 = 19/72

Finally, take the reciprocal to find R_total.

R_total = 72/19 ≈ 3.789 Ω

Ans. 3.789 Ω

Answer

We are given:

• Resistance = R = 10 Ω
• Voltage = V = 120 V

We want to find the power P consumed by the device.

We can use the formula:

P = V²/R

Plugging in the given values:

P = (120 V)²/10 Ω = 14400 V²/10 Ω = 1440 W

Ans. 1440 W

Answer

We are given:

• Power = P = 60 W
• Time = t = 5 hours

We want to find the energy E consumed by the light bulb.

The formula for energy is:

E = P × t

We can directly use the values in watts and hours to get the energy in watt-hours (Wh):

E = 60 W × 5 hours = 300 Wh

Answer

We are given:

• Number of cells = n = 100
• e.m.f of each cell = ε = 0.2 V

We need to find the resultant e.m.f when the cells are connected in series and in parallel.

(i) Series Connection:

When cells are connected in series, their e.m.fs add up. So, the total e.m.f (ε_series) is:

ε_series = n × ε = 100 × 0.2 V = 20 V

(ii) Parallel Connection:

When identical cells are connected in parallel, the total e.m.f remains the same as the e.m.f of a single cell, provided that all cells have the same e.m.f. So, the total e.m.f (ε_parallel) is:

ε_parallel = ε = 0.2 V

Answer

Given:

• R₁ = 4 Ω (in series with the entire network)
• R₂ = 6 Ω (upper branch of the parallel section)
• R₃ = 2 Ω, R₄ = 3 Ω (connected in series together in the lower branch)

Calculation:

Combining R₃ and R₄ (Series)

Since R₃ and R₄ in series.

R₃₄ = R₃ + R₄ = 3 Ω + 2 Ω = 5 Ω

Combining R₂ and R₃₄ (Parallel)

Now, R₂ = 6 Ω and R₃₄ = 5 Ω are in parallel.

1/R₂₃₄ = 1/R₂ + 1/R₃₄ = 1/6 + 1/5

1/R₂₃₄ = (5 + 6)/30 = 11/30, so R₂₃₄ = 30/11 = 2.73 Ω

Now, adding R₁ = 4 Ω in series with R₂₃₄. Now, the total resistance is:

R_total = R₁ + R₂₃₄ = 4 Ω + 2.73 Ω = 6.73 Ω

Therefore, the total resistance is 6.73 Ω.

Answer

Given data:

Resistance = R = 50 Ω; Voltage = V = 12 V; Time = t = 5 h

Required:

1. Power = P = ?
2. Current = I = ?
3. Energy = E = ?

Calculation:

1. Power = P = V²/R = (120)²/50 = 14400/50 = 228 W

2. Current = I = V/R = 120 V/50 Ω = 2.4 A

3. Energy = E = P × t = 228 W × 5 h = 1440 Wh

Ans. 288 W, 2.4 A, 1440 Wh

So, the results are:

1. Power consumed: 288 W
2. Current flowing through: 2.4 A
3. Electrical energy consumed (in 5 hours): 1440 Wh

(Ans. 10.75 Ω)

Answer

Given data:

Resistance in the circuit:

R₁ = 12.5 Ω; R₂ = 4.5 Ω; R₃ = 12.5 Ω

Required:

Total Resistance = R_total = ?

Calculation:

R₁ = 12.5 Ω and R₃ = 12.5 Ω are in parallel.

R₂ = 4.5 Ω is in series with the parallel combination of R₁ and R₃.

Now for parallel combination,

$$\frac{1}{R_{13}}=\frac{1}{R_1}+\frac{1}{R_3}=\frac{1}{12.5}+\frac{1}{12.5}=0.08+0.08=0.16$$

$$R_{13}=\frac{1}{0.16}=6.25\ \Omega$$

Now, calculating the total resistance R_total.

R₁₃ = 6.25 Ω and R₂ = 4.5 Ω are in series.

$$R_{\text{total}}=R_{13}+R_2=6.25+4.5=10.75\ \Omega$$

Therefore, the total resistance of the circuit is 10.75 Ω.