Question

electric heater of resistance 8 ohm draws 15 on from the service mains in 2 hours calculate the rate at which heat is developed in the heater​

Answer 1

R= 8ohm , I= 15A , T= 2hr

V= IR. (Ohm's law)

V= 15×8

V= 120volts

Now,

P= VI

P= 120×15

P= 1800 watt (W)

Used for 2 hour for 30 days

= 1800×2/1000

= 3.6 × 30

= 108 kWh

To find cost for rupees 5 per unit

= 108×5

= ₹540

Answer 2

R= 8ohm , I= 15A , T= 2hr

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law)

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now,

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days = 1800×2/1000

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days = 1800×2/1000 = 3.6 × 30

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days = 1800×2/1000 = 3.6 × 30 = 108 kWh

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days = 1800×2/1000 = 3.6 × 30 = 108 kWhTo find cost for rupees 5 per unit

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days = 1800×2/1000 = 3.6 × 30 = 108 kWhTo find cost for rupees 5 per unit = 108×5

R= 8ohm , I= 15A , T= 2hr V= IR. (Ohm's law) V= 15×8 V= 120volts Now, P= VI P= 120×15 P= 1800 watt (W)Used for 2 hour for 30 days = 1800×2/1000 = 3.6 × 30 = 108 kWhTo find cost for rupees 5 per unit = 108×5 = ₹540