Advertisements
Advertisements
प्रश्न
An electric bulb is marked 100 W, 250 V. How much current will the bulb draw if connected to a 250 V supply?
Advertisements
उत्तर
Given 250 V, I = ?
∵ P = VI
⇒ 100 = 250 × I
∴ I = `100/250 = 0.4` Amp.
संबंधित प्रश्न
Which unit could be used to measure current?
(a) Watt
(b) Coulomb
(c) Volt
(d) Ampere
The circuit diagram given below shows the combination of three resistors R1, R2 and R3:
Find :
(i) total resistance of the circuit.
(ii) total current flowing in the circuit.
(iii) the potential difference across R1.
In the circuit shown below, the voltmeter reads 10 V.

(a) What is the combined resistance?
(b) What current flows?
(c) What is the p.d. across 2 Ω resistor?
(d) What is the p.d. across 3 Ω resistor?
Three resistors of 6.0 𝛀, 2.0𝛀 and 4.0𝛀 are joined to an ammeter A and a cell of e.m.f. 6.0 V as
shown in fig 8.52 Calculate:
(a) the effective resistance of the circuit and
(b) the reading of ammeter.
Match the pairs.
| ‘A’ Group | ‘B’ Group |
| 1. Free electrons | a. V/R |
| 2. Current | b. Increases the resistance in the circuit |
| 3. Resistivity | c. Weakly attached |
| 4. Resistances in series | d. VA/LI |
Two resistors of 4 Ω and 6 Ω are connected in parallel. The combination is
connected across a 6 volt battery of negligible resistance. Calculate:
( i) The power supplied by the battery,
(ii) The power dissipated in each resistor.
Explain with the help of a diagram how does 'short circuiting' occur in an electric kettle.
The flow of electric charge per unit time is called potential difference.
