Electricity Worksheet6

Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?

Suppose a 6volt battery is connected across a lamp whose resistance is 20 Ω through a variable resistor as shown in the figure. If the current in the circuit is 0.25 A, calculate the value of the resistance from the resistor which must be used?

In the circuit diagram given in fig. (a), suppose the resistors R_{1}, R_{2} and R_{3} have values 5 Ω, 10 Ω and 30 Ω respectively, which have been connected to a battery of 12 V. calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

In given figure R_{1} = 10 Ω, R_{2} = 40 Ω and R_{3} = 3 Ω, R_{3} = 20 Ω, R_{5 }= 60 Ω and and a 12 V battery is connected to the arrangement, calculate (a) the total resistance and (b) the total current flowing in the circuit.

When two resistors of resistances R_{1} and R_{2} are connected in parallel, the net resistance is 3 Ω. When connected in series, its value is 16 Ω. Calculate the values of R_{1 }and R_{2}.
Answer:

R = 26 Ω,
As

Supply voltage, V = 6 volt
Resistance of the lamp, r = 20 Ω
Current in the circuit, I = 0.25 A
Let R_{1} be the resistance required from the variable resistance to be placed in series with the lamp.
Total resistance in the circuit, R = R_{1} + r = R_{1} + 20
Ohm’s law
Or
or

(a) since R_{1}, R_{2} and R_{3} are in parallel, the pd (V) across each one of them is the same, i.e., 12 V.
Thus, I_{1} (current through R_{1})
I_{2} (current through R_{2})
I_{3} (current through R_{3})
(b) total current in the circuit,
I = I_{1} + I_{2} + I_{3} = 2.4 + 1.2 + 0.4 A
(c) If R_{P} is the total resistance in the circuit, then
or R_{p} = 3 Ω
we can also find the total current by first calculating R_{P} and then using,

since R_{1} and R_{2} are in parallel, their resultant resistance (R_{P}) is given by
Or
Further, as R_{3}, R_{4} and R_{5} are in parallel, their resultant resistance (R_{P}) is given by
R_{P} = 10 Ω
As R_{P} and R_{p} are in series, total resistance in the circuit,
R_{s} = R_{p} + R_{p} = 8 + 10 = 18 Ω
Total current in the circuit, i.e.,

When R_{1} and R_{2} are connected in parallel, net resistance (R_{P}) is given by
Or …(i)
When R_{1} and R_{2} are connected in series, net resistance (R_{s}) given by
R_{s} = R_{1} + R_{2} = 16 ….(ii)
From (i) and (ii),
Or R_{1}(16 – R_{1}) = 48
(as R_{1} + R_{2} = 16, R_{2} = 16 – R_{1})
Or 26R_{1} – R_{1}^{2} = 48
Or R_{1}^{2} – 16R_{1} + 48 = 0
Or (R_{1} – 12) (R_{1} – 4) = 0
Thus, either R_{1} = 12 Ω or 4 Ω
R_{2} = 4 Ω or 12 Ω
Therefore, the resistances of two resistors are 4 Ω and 12 Ω.