For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm�s law asV= IR
R= V/I
Where,
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of the combination = R,given as



Therefore, four resistors of 176 Ω are required to draw the given amount of current.

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2 = 3 Ω is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel



Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be



The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.
(b) Two resistor in series



Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω.

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be



Therefore, the total resistance is 4 Ω.

Resistance R1 of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts
Because R=V²/P



∴ Number of electric bulbs connected in parallel are 110.

Supply voltage, V= 220 V
Resistance of one coil, R= 24 Ω



(i) Coils are used separately

According to Ohm�s law,
V= I₁R₁
Where,

I1 is the current flowing through the coil

I1 = V/R₁ = 220/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, R₂ = 24 Ω + 24 Ω = 48 Ω

According to Ohm�s law,V = I₂R₂
Where,

I₂ is the current flowing through the series circuit

I₂ = V/R₂ = 220/48 = 4.58 A
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel

Total resistance, R₃ is given as =



According to Ohm�s law,

V= I₃R₃
Where,
I₃ is the current flowing through the circuit I3 = V/R₃ = 220/1₂ = 18.33 A
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

(i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
I= 6/3 = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression,
P= (I)2R = (2)2 x 2 = 8 W

(ii) Potential difference, V = 4 V
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V.
Power consumed by 2 Ω resistor is given by
P= V2/R = 42/2 = 8 W
Therefore, the power used by 2 Ω resistor is 8 W.

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,Power = Voltage x Current
Current =  Power/Voltage = 60/220 A
Hence, current drawn from the line = 100/220 + 60/220 = 0.727 A

Energy consumed by an electrical appliance is given by the expression,H= Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 ×3600 = 9 ×10⁵ J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
= 7.2×10⁵ J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Rate of heat produced by a device is given by the expression for power as, P= I²R
Where,
Resistance of the electric heater, R= 8 Ω
Current drawn, I = 15 A

P= (15)² x 8 = 1800 J/s

Therefore, heat is produced by the heater at the rate of 1800 J/s.