5 bulbs are connected in series in a room. One bulb is fused. It is removed and remaining 4 bulbs are again connected in series to the same circuit. What will be the effect on the following physical quantities? (Increases, Decreases, Remain Same)
(a) Resistance
(b) Intensity of light
(a) In a series connection, total resistance is the sum of individual resistances.
Let resistance of each bulb be R.
Initially :
Rtotal = 5R
After removing one bulb:
Rtotal = 4R
So, resistance will decrease.
(b) When resistance decreases, total current in the circuit (assuming the same voltage supply) will increase :
I = V/R
So, more current → more power delivered to each bulb.
Since,
Power (P) = I2R,
Hence, intensity of light increases for each remaining bulb.
More ICSE Easy level questions
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A resistance R is connected across a cell with a switch and a rheostat in series. A voltmeter is connected parallel across the cell. Current in the circuit is increased using the rheostat. (a) How will the voltmeter reading change? (increase / decrease / remain the same) (b) Justify your answer stated in (a) above.
A current flows through a metallic conductor for a long period of time. State the change you would expect in its: (a) Resistance (b) Resistivity
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Find the value of current I drawn from the cell. (a) Calculate the current I. (b) Calculate the terminal voltage.
Three identical bulbs B1, B2 and B3, each of power rating 18 W, 12 V, are connected to a battery of 12 V as shown. (a) Calculate: (i) the resistance of each bulb (ii) the current drawn from the cell (b) If bulb B3 is removed from the circuit, will the brightness of bulb B1 increase, decrease or remain the same?
In a closed circuit containing a bulb and a cell, the electromotive force (ε) and the terminal voltage (V) is related as. (Given I is current and r is internal resistance.) (a) V = ε + Ir (b) V = ε – Ir (c) V = ε ÷ Ir (d) V = ε × Ir
Choose the correct relation for e.m.f. (ε) and terminal voltage V: (a) ε = V [always] (b) V > ε [always] (c) V < ε [when the cell is in use] (d) None of these
