Electricity Worksheet-6

Electricity Worksheet-6

1. 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?

2. Suppose a 6-volt 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?

3. In the circuit diagram given in fig. (a), suppose the resistors R₁, R₂ and R₃ 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.

4. In given figure R₁ = 10 Ω, R₂ = 40 Ω and R₃ = 3 Ω, R₃ = 20 Ω, R₅  = 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.

5. When two resistors of resistances R₁ and R₂ are connected in parallel, the net resistance is 3 Ω. When connected in series, its value is 16 Ω. Calculate the values of R₁ and R₂.

Answer:

1. R = 26 Ω,

As

2. Supply voltage, V = 6 volt

Resistance of the lamp,     r = 20 Ω

Current in the circuit,        I = 0.25 A

Let R₁ be the resistance required from the variable resistance to be placed in series with the lamp.

Total resistance in the circuit, R = R₁ + r = R₁ + 20

Ohm’s law

Or    

or     

3. (a) since R₁, R₂ and R₃ are in parallel, the pd (V) across each one of them is the same, i.e., 12 V.

Thus,         I₁ (current through R₁)

                   I₂ (current through R₂)

                   I₃ (current through R₃)

(b)    total current in the circuit,

          I = I₁ + I₂ + I₃ = 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,

4. since R₁ and R₂ are in parallel, their resultant resistance (R_P) is given by

Or    

Further, as R₃, R₄ and R₅ 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.,

5. When R₁ and R₂ are connected in parallel, net resistance (R_P) is given by

Or                …(i)

When R₁ and R₂ are connected in series, net resistance (R_s) given by

          R_s = R₁ + R₂ = 16       ….(ii)

From  (i) and (ii),

Or     R₁(16 – R₁) = 48                          

          (as R₁ + R₂ = 16, R₂ = 16 – R₁)

Or     26R₁ – R₁² = 48

Or     R₁² – 16R₁ + 48 = 0

Or     (R₁ – 12) (R₁ – 4) = 0

Thus, either       R₁ = 12 Ω or 4 Ω

                             R₂ = 4 Ω or 12 Ω

Therefore, the resistances of two resistors are 4 Ω and 12 Ω.