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 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?
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.
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.
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.
R = 26 Ω,
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
(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 RP is the total resistance in the circuit, then
or Rp = 3 Ω
we can also find the total current by first calculating RP and then using,
since R1 and R2 are in parallel, their resultant resistance (RP) is given by
Further, as R3, R4 and R5 are in parallel, their resultant resistance (RP) is given by
RP = 10 Ω
As RP and Rp are in series, total resistance in the circuit,
Rs = Rp + Rp = 8 + 10 = 18 Ω
Total current in the circuit, i.e.,
When R1 and R2 are connected in parallel, net resistance (RP) is given by
When R1 and R2 are connected in series, net resistance (Rs) given by
Rs = 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.