Question

Solve the following:

(1) Complete the following

(a)

(b)

(2) Find the total resistance in the circuit.

(3) Find the total resistance and current in the circuit.

(4) If two resistors are connected in series, the total resistance is 45Ω and if the same resistors are connected in parallel the total resistance becomes 10Ω. Find the value of individual resistors.

(5) If a bulb of 60W is connected across a source of 220 V, find the current drawn by it.

(6) A potential difference of 250 Volts is applied across a resistance 1000 Ω in an electric iron. Find (1) the current and (2) heat energy produced in Joules in 12 sec.

(7) A washing machine rated 300W is operated for one hour/day. If the cost of unit is Rs 3.00 find the cost of the energy to operate a washing machine for the month of March.

(8) If a TV of rating 100W is operated for 6 hrs per day, find the number of units consumed in any leap year.

(9) How will you connect three resistances of 4 ohms each to get 12 Ω, 6 Ω and Ω, respectively.

Answer 1

(1)

(a) We can see that all the resistors are connected in a simple series combination.

∴ 20Ω + 3.5Ω + 4.9Ω = 28.4Ω

(b) In this case all the resistors are connected in a simple parallel combination,

Ω

Let us see how.

(2) Total resistance, R = ?

As resistances 8Ω and 12Ω are in parallel combination and then connected to another resistance 7.2Ω in series,

First let us consider parallel section of the circuit and find equivalent resistance. For that,

Now taking the series combination of this R_p and 7.2Ω into account,

Total resistance, R = R_p+ 7.2

(3) Case I: Total resistance, R = ?

As resistances 3Ω and 2Ω are in series combination and then connected to another resistance 5Ω by parallel means,

R_s = 3Ω+2Ω = 5Ω

For total resistance,

Case II: Current, I = ?

From Ohm’s law, I = V/R

V = 4 V (given)

I = 4 / 2.5

∴ I = 1.6 A

(4) Let the two resistors be R₁ and R₂.

According to question,

In parallel combination, R₁R₂/(R₁+R₂) = 10Ω ……………(1)

In series combination, R₁+R₂ = 45Ω …………………. (2)

Putting the value of (R₁+R₂) in equation (1) from equation (2), we get

R₁R₂ = 10×45 = 450Ω

Now,

From the identity, (R₁- R₂)² = (R₁+R₂)² − 4R₁R₂

We get R₁- R₂ = 15Ω …………… (3)

Add equations (2) and (3),

2R₁ = 60 Ω

⇒ R₁ = 30 Ω

Plug-in the value of R₁ in equation (3),

R₂ = (30 − 15) = 15Ω

(5) Given:

Power, P = 60W

Potential difference, V = 220 V

Current, I = ?

We know, P = VI

∴ I = P/V

I = 60/220 = 0.2727A

(6) Given:

Potential difference, V = 250 V

Resistance, R = 1000Ω

Case I: Current, I = ?

From Ohm’s law, V = IR

∴ I = V/R

I = 250/1000 = 0.25A

Case II: Heat energy produced in 12 sec in joules, H = ?

From Joule’s law, H = I²Rt

Putting values, we get

H = 0.25×0.25×1000×12 = 750J

(7) Here, washing machine rated 300W is operated for one hour a day. We have to calculate the electricity bill for the month of March which means machine had been operated for 31 days or 31 hours to be precise.

First we need to calculate the net energy usage in kWh which is the commercial unit of electricity.

We know, Power, P = E/t

Units consumed in one day = 300×1Wh = 0.3 KWH = 0.3 unit

Total units consumed in the month of March = 0.3× 31 units = 9.3 units

Now, the electric bill is at the cost of Rs.3.00 per unit.

So, total cost for the month of March = 9.3 × 3 = Rs. 27.90

(8) Given: Power of TV, P = 100W.

Time of operation, t = 6 hours/day.

Number of energy units consumed in a leap year = ?

We know, P = E/t

Number of days in a leap year = 366 days

Now total time of operation for TV in a leap year = 6 × 366 hours

Thus, energy consumed by TV in a leap year, E = P×t

⇒ E = 100×6×366 WH

⇒ E = 219600 WH

⇒ E = 219.6 KWH = 219.6 units

(1 commercial unit = 1 kWh)

(9) We have given three resistors of resistance 4Ω each.

Case I: To get an equivalent resistance, R = 12Ω,

We will connect all the given resistors in a series combination since the equivalent resistance of the resistors connected in series is given by the algebraic sum of their individual resistances which will give us the desired result.

Case II: To get an equivalent resistance, R = 6Ω,

First we will connect two resistors in parallel manner which will make the effective resistance 2Ω and then setting a series combination between this effective resistance and third resistor of 4Ω resistance will serve our purpose. See the steps below:

(i)

(ii)

Case III: To get an equivalent resistance, R = 4/3Ω,

We will connect all the given resistors in parallel combination since in parallel combination equivalent resistance is given as:

Which will give us the desired result of 4/3 Ω.