The heating coil of an electric heater has two resistance coils I and II, each of resistance 36 ,
which may be used in series or in parallel. When the heater is a connected to a 220 V supply line,
what will be the currents drawn in each case?
Answers
Answer:
Case I: 9.17 Amperes
Case II: 4.58 Amperes
Case III: 18.33 Amperes
Step-by-step explanation:
Given:
Voltage(V) = 220 Volts
Resistance of Coil A, Ra = 24 ohm
Resistance of Coil B, Rb = 24 ohm
Case I: When coils are used separately.
Applying the formula,
\fbox{V = IR}V = IR
We get,
I = \frac{V}{R}I=RV
I = \frac{220}{24} = 9.17 AI=24220=9.17A
Therefore, the current flowing in this circuit is 9.17 Amperes.
Case II: When coils are used in series.
Equivalent Resistance,
R = 24 + 24 = 48 ohms
Applying Formula V = IR,
I = \frac{V}{R}I=RV
I = \frac{220}{48} = 4.58AI=48220=4.58A
Therefore the current flowing is 4.58 Amperes.
Case III: When coils are used in parallel.
Equivalent Resistance,
R = \frac{R1R2}{R1 + R2}R=R1+R2R1R2
R = \frac{24 * 24}{24 + 24}R=24+2424∗24
R = \frac{576}{48} = 12 OhmsR=48576=12Ohms
Current Flowing,
I = \frac{V}{R}I=RV
I = \frac{220}{12} = 18.33AI=12220=18.33A
Therefore, the current flowing is 18.33 Amperes.
Given V=220 V
RA = RB = 24 ohm
(a) Current drawn when only coil A is used:
I = V/RA = 220/24
=9.16 amps
(b) Current drawn when coils A and B are used in series:
Total resistance, R = RA + RB = 24 + 24 = 48 ohms
I = V/R = 220/48
= 4.58 amps
(c) Current drawn when coils A and B are used in parallel:
Total resistance, 1/R = 1/RA + 1/RB = 1/24 + 1/24 = 2/24 = 1/12
R=12 ohms
I = V/R = 220/12
=18.33 amps