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Electricity and Magnetism (Numerical Problem)

Q. An electric iron of 1000w is connected in 220v in main line, what should be the capacity of the fuse? Soln→ Power (P) = 1000w volt (V) = 220v current (I) = ? By the formula P = I V or, I = $\frac{P}{V}$ or, I = $\frac{\mathrm{1000}}{\mathrm{220v}}$ or, I = 4.54 A Therefor, suitable fuse is 5 A for this electric heater. Q. What will be the capacity of a fuse to operate a heater of 0.8 kw with 220v main supplies? Q. If an iron of 1000w, two bulbs of each 100w, a television of 60 w and a radio of 40w area used in a house, wht should be the capacity of fuse used? the voltage of the supplyied line is 220v.?	Q. A domestic wiring has a fuse of 5A. What is the maximum number of 100watt bulb that can be used in the supply of 220v? Soln→ Maximum current (I) = 5A volt (V) = 220v power of bulb (p) = 100w Suppose maximum number of bulb can be connected is 'n' By the formula, P = I V or n.p = 5 × 220 or, n × 100 = 1100 or, n = $\frac{\mathrm{1100}}{\mathrm{100}}$ or, n = 11 Therefor, 11 bulbs can be connected. Q. What is the the amount of current needed for a fluorescent lamp of 40w connected to a source of 220v?
Q. Ten electric bulbs of 100 watt each and two electric heaters of 1000 watt each are used for 6 hours continuously. Calculate the unit of electricity consumed? Soln→ Number of bulbs n = 10 Power of bulb (P) = 100w = $\frac{\mathrm{100}}{\mathrm{1000}}$kw                        = 0.1kw time (t) = 6 hr. Electricity consumed = P n t                   = 0.1 × 10 × 6 = 6 unit Number of heater n = 2 Power of heater (P) = 1000w = $\frac{\mathrm{1000}}{\mathrm{1000}}$kw                            = 1kw time (t) = 6 hr. Electricity consumed = P n t                       = 1 × 10 × 6 = 60 unit Hence total electricity consumed = 6 + 60 = 66 unit. Q. An electric kettle rated 220v and 2.2kw works for 3 hours. Find the energy consumed and the current drawn by it?	Q. Four electric bulbs of 60w power each and a heater of 1000w power are used in a home for 4 hours daily respectively. Calculate the monthly cost of electricity at the rate of Rs. 7.50 per unit. Claculate the main current when all these appliances run together. (Take the voltage supply is 220v). Soln→ Number of bulbs n = 4 Power of bulb (P) = 60w = $\frac{\mathrm{60}}{\mathrm{1000}}$kw                        = 0.06kw time (t) = 4 hr. Electricity consumed = P n t                   = 0.06 × 4 × 4 = 0.96 unit Number of heater n = 1 Power of heater (P) = 1000w = $\frac{\mathrm{1000}}{\mathrm{1000}}$kw                            = 1kw time (t) = 4 hr. Electricity consumed = P n t                       = 1 × 1 × 4 = 4 unit Hence total electricity consumed = 0.96 + 4 = 4.96 unit. Rate per unit is Rs. 7.50 Montly cost = (4.96 × 7.50) × 30 = Rs 1116. Q. Ten bulbs of 100 watt each are used for 6 hrs and 3 electric heaters of 2kw each are used for 2hrs. Calculate the total consumption of electricity for a day.
Q. The number of turns in the primary winding of trnsformer is 200 times more than that in the secondary winding. Calculate the input emf in the primary winding if emf generated in the secondary winding in 220 volt AC. Soln→Number of turns in primary winding n1 = 200 k Number of turns in secondary winding n2 = k Seconadary volt v2 = 220v Primary volt v1 ? By the formula, $\frac{\mathrm{v<sub>1</sub>}}{\mathrm{n<sub>1</sub>}}$ = $\frac{\mathrm{v<sub>2</sub>}}{\mathrm{n<sub>2</sub>}}$                 or, $\frac{\mathrm{v<sub>1</sub>}}{\mathrm{200\; k}}$ = $\frac{\mathrm{220}}{k}$                 or, v1 = $\frac{\mathrm{220}}{k}$ × 200 k                 or, v1 = 44000 v Q. The number of turns in the primary winding of a certian transformer is 150 times more than that in the secondary winding. Calculate the input emf in the primary winding if the emf generated in the secondary winding in 220v AC.	Q. Primary voltage is 220v, what will be secondary voltage if ratio of number turns in secondary coil to primary coil is 1/4? Soln→ Primary voltage v1 = 200v Secondary voltage v2 = ? $\frac{\mathrm{secondary\; coil\; n<sub>2</sub>}}{\mathrm{primary\; coil\; n<sub>1</sub>}}$ = $\frac{1}{4}$ By the formula, $\frac{\mathrm{v<sub>2</sub>}}{\mathrm{v<sub>1</sub>}}$ = $\frac{\mathrm{n<sub>2</sub>}}{\mathrm{n<sub>1</sub>}}$                 or, $\frac{\mathrm{v<sub>2</sub>}}{\mathrm{200}}$ = $\frac{1}{4}$                 or, v2 = $\frac{1}{4}$ × 200                 or, v2 = 50v Q.To listen 12v radio in a main line 240v with transformer of primary turns 2000. What will be the secondary turns.
Q. ? Soln→	Q. ? Soln→