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CLASS - 10 SCIENCE IMPORTANT QUESTIONS OF MARK 5   FOR BOARD EXAMINATION

QUESTION NO. 17

An electric lamp of resistance 20 ohm and a conductor of resistance 4 ohm are connected to 6V battery as shown in the circuit.
(a) the total resistance of the circuit
(b) the current through the circuit
(c) the potential difference across the 
     (i) electric lamp
     (ii) conductor
(d) power of the lamp.

ANSWER NO. 17


(a) Voltage of battery V = 6 volt
Resistance of electric lamp R1 = 20 ohm
Resistance of conductor   = 4 ohm
 Total resistance  R = R1 + R2
                                     = 20 + 4 = 24 ohm
(b) According to ohm's law = V = IR
                                                I = V/R
                                                  = 6/24 = 0.25 ampere
(c) (i) Potential difference across the electric lamp.
    V = IXR1 = 0.25 X 20 = 5 volt.
(ii) Potential difference across the conductor
   V = IXR2 = 0.25 X 4 = 1 volt
(d) Power of lamp = P = VI
                                         = 5 X 0.25 = 1.25 watt.

QUESTION NO. 18


(a) Three resistors of resistances R, R, R are connected.
    (i) in series and
    (ii) in parallel
  write expression for the equivalent ree of the combination in each case.
(b) Two identical  resistors of 12 ohm each are connected to a battery of 3V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.

ANSWER NO. 18


(a) (i) Equivalent resistance R = R1 + R2 + R3
(ii) Equivalent resistance 1/R = 1/R1 +1/R2 + 1/R3
(b) Since resistance are identical of 12 ohm so 
   R = R1 + R2
       = 12 + 12 = 24 ohm
Applied voltage is given V = 3V
  P1 = vv/R = 3X3/2X2 = 9/24 W
If connected in parallel 
1/R = 1/R1 + 1/R2 = 1/12 + 1/12 = 2/12 = 1/6
      R = 6 ohm
 P2 = vv/R = 3X3/6 = 9/6 W
Ratio of power consumed in the two combination is.

     P1/P2 = 9/24  =1/4
              9/6
   P1 : P2 = 1 : 4

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