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Answer :
Supply voltage, V= 220 V
Resistance of one coil, R= 24 \(\Omega \)
(i) Coils are used separately
According to Ohm's law,
\( V= I_1R_1 \)
Where,
\( I_1 \) is the current flowing through the coil
\( I_1 = \frac{V}{R_1} = \frac{220}{24} = 9.166 A \)
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, \(R_2 \) = 24 \(\Omega \) + 24 \(\Omega \) = 48 \(\Omega \)
According to Ohm's law,V = \( I_2R_2 \)
Where,
\( I_2 \) is the current flowing through the series circuit
\( I_2 = \frac{V}{R_2} = \frac{220}{48} = 4.58 A \)
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 \( = \frac{24}{2} = 12V \)
According to Ohm's law,
\( V= I_3R_3 \)
Where,
I3 is the current flowing through the circuit \( I_3 = \frac{V}{R_3} = \frac{220}{12}= 18.33 A \)
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.