Question

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And need solutions for all question​

Answer 1

Answer:

ANSWER

1st QUESTION ANSWER

a.) This means that if it is kept 150 volts in its presence, then it will consume 60 more electricity in 1 second.

b.)

$i \: = \frac{p}{v} = \frac{60w}{150v} = 0.4$

2nd QUESTION ANSWER

a.)

$p = 1.5 \: kwt = 30minute \\ \\ \frac{30}{60} = \frac{1}{2} hour \\ \\ 1.5 \times \frac{1}{2} = 0.75kwh$

b.)

$1 \: kwh \: = 3.6 \times {10}^{6}j \\ \\ 75 \: kwh \: = 0.75 \times 3.6 \times {10}^{6} \\ \\ 2.700 \times {10}^{6} \\ \\ 2.7 \times {10}^{6}j$

3rd QUESTION ANSWER

$\fbox\red{electric \: energy \: consumed \: per \: day} \\ = 12 \: kwh \\ \\ \fbox\blue{in \: 30 \: days} = 12 \times 30 = 360 \\ \\ \fbox\green{unit \: cost \: paid \: in \: 1 \: month} = 360 \times 6.25 = 2250$

4th QUESTION ANSWER

a.)

$\fbox\red{5 \: bulbs} = 5 \times 100 = 500w \\ \\ \fbox\pink{2 \: fans} = 2 \times 60 = 120w \\ \\ \fbox\green{2 \: a.c \:} = 2 \times 1.5 \times 1000 = 3000w \\ \\ \fbox\blue{total \: power} = 3620w$

b.)

$\fbox\red{in \: 30 \: days} = \frac{3620}{1000} \times 30 = 108.6 \: kw$

c.)

$\fbox\green{total \: electrical \: energy \: consumed \: in \: 30 \: days} \\ \\ = t \: = 5 \: hours \\ \\ q = p \: t \\ \\ q \: = 108.6 \times 5 \\ \\ = 543 \: kwh$

d.)

$\fbox\red{cost} = 543 \times 6.25 = 3393.75$

I hope it helps you didu

# silent girl answer

Answer 2

C. Numerical

1. An electrical appliance is rated as 60 W - 150V

• a) What do you understand by this statement?
• (b) How much current will flow through the appliance when in use ?

Answer:

• a) It means that if the potential difference of 150 V is applied through the circuit, the power consumed is 60 W or in other words the appliance will consume 60 W electrical power.

• b) Electric power P = VI
• 60 W = 150 V × I
• I = 60/150
• I = 6/15
• I = 0.4 A

2. An electric iron of power 1.5 kW is used for 30 minute to press the clothes. Calculate the electrical energy consumed in (a) kilowatt hour (b) joule.

Answer:

a)

Here, power of an electric iron, P = 1.5 kW and the time of use, t = 30 minutes = 1/2 h

❖ Electrical energy, E = P × t

➙ E = 1.5 kW × 1/2 h

➙ E = 15/10 × 1/2 h

➙ E = 0.75 kW h

b)

❖ 1 kW h = 3.6 × 10⁶ J

➙ 75 KW h = 75 × 3.6 × 10⁶ J

= 2.700 × 10⁶

= 2.7 × 10⁶ J

3. Assuming the electric consumption per day to be 12 kWh and the rate of electricity to be ₹6.25 per unit, find how much money is to be paid in a month of 30 days ?

Answer:

The electric energy consumed per day, E = 12 kWh

Total electrical energy consumed in 1 month of 30 days

= nE = 30 × 12 kW h

= 360 kW h

= 360 units

Now,

⟹ Cost to be paid in 1 month = Number of units × rate

⟹ Cost to be paid in 1 month = 360 × ₹6.25

⟹ Cost to be paid in 1 month = ₹2250

4. In a premise 5 bulbs each of 100 W, 2 fans each of 60 W, 2 A.Cs each of 1.5 kW are used for 5 h per day. Find

• (a) total power consumed per day,
• (b) total power consumed in 30 days,
• (c) total electrical energy consumed in 30 days,
• (d) the cost of electricity at the rate of ₹6.25

Answer:

⟹ Power consumed by 5 bulbs = 5 × 100

⟹ Power consumed by 5 bulbs = 500 W

⟹ Power consumed by 2 fans = 60 × 2

⟹ Power consumed by 2 fans = 120 W

⟹ Power consumed by 2 A.Cs = 1.5 × 2

⟹ Power consumed by 2 A.Cs = 3 kW

⟹ Power consumed by 2 A.Cs = 3 × 1000 W [ 1 kW = 1000 W = 10³ W ]

⟹ Power consumed by 2 A.Cs = 3000 W

a)

➤ Total power consumed per day = Power consumed by 5 bulbs + Power consumed by 2 fans + Power consumed by 2 A.Cs

➤ Total power consumed per day = 500 + 120 + 3000

➤ Total power consumed per day = 3620 W

b)

➤ Total power consumed per day = 3620 W

➤ Total power consumed in 30 days = 3620 × 30

➤ Total power consumed in 30 days = 108600 W

➤ Total power consumed in 30 days = 108600/1000 [ 1 kW = 1000 W = 10³ W ]

➤ Total power consumed in 30 days = 1086/10

➤ Total power consumed in 30 days = 108.6 kW

c)

Total electric energy is used for 5 h per day

➤ Total electrical energy consumed in 30 days, E = P × t

➤ Total electrical energy consumed in 30 days, E = 108.6 kW × 5 h

➤ Total electrical energy consumed in 30 days = 543 kW h

d)

➤ The cost of electricity = Number of units × rate

➤ The cost of electricity = 543 × ₹6.25

➤ The cost of electricity = ₹3393.75

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