in coil B, = - 3. = 500 × = -500 × = -200 V M = = = 0.5H # A circuit consists of 2 parallel resistors having resistances 20Ω and 30Ω respectively, connected in series with a 15Ω resistor. If the current through 30Ω resistor is 1.2A, Find Currents in 20Ω and 15Ω resistors (ii) The voltage across the whole circuit (iii) voltage across 15 resistor and 20 resistor (iv) total power consumed in the circuit. Ω Ω [Dec 2014/Jan2015] Solution : Voltage across 30 is = I R = 1.2 × 30 = 36V Ω Current in 20 is = V/R = 36/ 20 = 1.8 A (Since 20 and 30 are in parallel) Ω Ω Ω Total current in the circuit is = 1.8 +1.2 = 3 A Voltage in 15 is = 3 ×15 = 45 V Ω Total voltage is = 45 +36 = 81 V in the circuit is = VI = 81 ×3 = 243 W # A coil consists of 600 turns and a current of 10A in the coil gives rise to a magnetic flux of 1mWb.Calculate (i) self-inductance (ii) induced emf (iii) energy stored when the current is reversed in 0.01second. [Dec 2014/Jan2015] L = = 0.06 = H = - = e = -L 0.06 (-10-10)/0.01 120V = Energy stored = (1/2)* (0.06)* 102 = 3 J
3. in coil B, = = - 500 × = -500 × = -200 V M = = = 0.5H # A circuit consists of 2 parallel resistors having resistances 20Ω and 30Ω respectively, connected in series with a 15Ω resistor. If the current through 30Ω resistor is 1.2A, Find Currents in 20Ω and 15Ω resistors (ii) The voltage across the whole circuit (iii) voltage across 15 Ω resistor and 20 Ω resistor (iv) total power consumed in the circuit. [Dec 2014/Jan2015] Solution : Voltage across 30 Ω is = I R = 1.2 × 30 = 36V Current in 20 Ω is = V/R = 36/ 20 = 1.8 A (Since 20 Ω and 30 Ω are in parallel) Total current in the circuit is = 1.8 +1.2 = 3 A Voltage in 15 Ω is = 3 ×15 = 45 V Total voltage is = 45 +36 = 81 V in the circuit is = VI = 81 ×3 = 243 W # A coil consists of 600 turns and a current of 10A in the coil gives rise to a magnetic flux of 1mWb.Calculate (i) self-inductance (ii) induced emf (iii) energy stored when the current is reversed in 0.01second. [Dec 2014/Jan2015] L = = = 0.06 H e = -L = - 0.06 (-10-10)/0.01 = 120V Energy stored = = (1/2)* (0.06)* 102 = 3 J
Module-I.Lec-10Numerical-examples-on-Dc-cituits-and-Magnetism.docx