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Alternating current converted
- 1. Alternating current The current or emf whose magnitude changes with time and direction reverses periodically is called an alternating current or emf.
- 2. Mean or average value of alternating current The average or mean value of 𝑎. 𝑐. over a complete cycle is zero. So it is defined in half cycle of 𝑎. 𝑐. It is defined as ‘the steady value of current which sends the same amount of charge through the circuit in half cycle of 𝑎. 𝑐. as it is done by 𝑎. 𝑐. in same time in same circuit’. It is denoted by 𝐼𝑚 𝑜𝑟 𝐼𝑎𝑣. The instantaneous value of current is given by 𝐼 = 𝐼0 sin 𝜔𝑡 If this current flows in the circuit then small amount of charge sent in small time dt is 𝑑𝑞 = 𝐼𝑑𝑡 = 𝐼0 sin 𝜔𝑡 𝑑𝑡 The total amount of charge supplied in half cycle of 𝑎. 𝑐. is obtained by integrating above equation from 𝑜 𝑡𝑜 𝑇 2 .
- 3. 𝑖𝑒, 𝑞 = 𝑜 Τ 𝑇 2 𝐼0 sin 𝜔𝑡 𝑑𝑡 = 𝐼0 𝑜 Τ 𝑇 2 sin 𝜔𝑡 𝑑𝑡 = 𝐼0 − cos 𝜔𝑡 𝜔 0 Τ 𝑇 2 = − 𝐼0 𝜔 cos 𝜔𝑡 0 Τ 𝑇 2 = − 𝐼0 𝜔 cos 𝜔𝑇 2 − cos 0 = − 𝐼0 𝜔 cos 𝜋 − cos 0 ∵ 𝜔 = 2𝜋 𝑇 = − 𝐼0 𝜔 −1 − 1 = 2𝐼0 𝜔 ………………….. (𝑖) If 𝐼𝑚 be the mean value of an 𝑎. 𝑐. over positive half cycle then the charge send in time T/2 is given by
- 4. 𝑞 = 𝐼𝑚 × 𝑇 2 … … … … … … . . (𝑖𝑖) Since, the charge in equation (𝑖) and (ii) is same 𝑖𝑒, 2𝐼0 𝜔 = 𝐼𝑚 × 𝑇 2 𝑜𝑟, 𝐼𝑚 = 4𝐼0 𝜔𝑇 𝑜𝑟, 𝐼𝑚 = 4𝐼0 2𝜋 ∵ 𝜔 = 2𝜋 𝑇 = 2 𝜋 𝐼0 = 0.637𝐼0 Hence, the average value of 𝑎. 𝑐. over positive half cycle is 0.637 times the peak value of 𝑎. 𝑐. or 63.7% of its peak value. Similarly, for negative half cycle, 𝐼𝑚 = −0.637𝐼0
- 5. Root mean square (R.M.S) value of A.C. It is defined as ‘ the steady value of current which when passed through a resistor in complete cycle of 𝑎. 𝑐. produces the same amount of heat as it is done by 𝑎. 𝑐. in same time in same resistor’. It is denoted by 𝐼𝑟𝑚𝑠. It is also known as virtual value or effective value of 𝑎. 𝑐.. The instantaneous value of current is given by 𝐼 = 𝐼0 sin 𝜔𝑡 If this current flows in resistor R then small amount of heat produced in small time dt is 𝑑𝐻 = 𝐼2 𝑅𝑑𝑡 = 𝐼0 sin 𝜔𝑡 2 𝑅𝑑𝑡 = 𝐼0 2 𝑅𝑠𝑖𝑛2𝜔𝑡𝑑𝑡 The total amount of heat is obtained by integrating above expression from o to T
- 6. 𝑖𝑒, 𝐻 = 0 𝑇 𝐼0 2 𝑅𝑠𝑖𝑛2𝜔𝑡𝑑𝑡 = 𝐼0 2 𝑅 0 𝑇 𝑠𝑖𝑛2𝜔𝑡𝑑𝑡 = 𝐼0 2 𝑅 0 𝑇 1−cos 2𝜔𝑡 2 𝑑𝑡 = 𝐼0 2𝑅 2 0 𝑇 1 − cos 2𝜔𝑡 𝑑𝑡 = 𝐼0 2𝑅 2 0 𝑇 𝑑𝑡 − 𝑜 𝑇 cos 2𝜔𝑡 𝑑𝑡 = 𝐼0 2𝑅 2 𝑡 0 𝑇 − sin 2𝜔𝑡 2𝜔 0 𝑇 = 𝐼0 2𝑅 2 𝑇 − 0 − 1 2𝜔 sin 2𝜔𝑇 − sin 0 = 𝐼0 2𝑅 2 𝑇 = 𝐼0 2𝑅𝑇 2 ……………. (𝑖)
- 7. If 𝐼𝑟𝑚𝑠 be the 𝑟. 𝑚. 𝑠 value of 𝑎. 𝑐., then the amount of heat produced in the same resistance R in same time T is 𝐻 = 𝐼𝑟𝑚𝑠 2 𝑅𝑇………. (ii) Since the heat in equation (𝑖) and (ii) are same, 𝑖𝑒, 𝐼𝑟𝑚𝑠 2 𝑅𝑇 = 𝐼0 2𝑅𝑇 2 𝑜𝑟, 𝐼𝑟𝑚𝑠 2 = 𝐼0 2 2 ∴ 𝐼𝑟𝑚𝑠 = ± 𝐼0 2 = ±0.707𝐼0 Hence, the RMS value of 𝑎. 𝑐. over complete cycle is 0.707 times the peak value of 𝑎. 𝑐. or 70.7% of its peak value. Similarly, for RMS value of emf 𝜀𝑟𝑚𝑠 = ± 𝜀0 2 = ±0.707𝜀0
- 8. A.C. through a Resistance Suppose a resister of resistance R is connected in series with 𝑎. 𝑐. source as shown in figure. 𝐼 is the current supplied by 𝑎. 𝑐. source. The instantaneous value of emf is given by 𝜀 = 𝜀0 sin 𝜔𝑡………. (𝑖) Now, the current flowing through the resistor is 𝐼 = 𝜀 𝑅 = 𝜀0 sin 𝜔𝑡 𝑅 ……….. (𝑖𝑖) For I to be maximum, sin 𝜔𝑡 = 1 𝐼0 = 𝜀0 𝑅 ………….. (𝑖𝑖𝑖) Therefore equation (ii) becomes, 𝐼 = 𝐼0 sin 𝜔𝑡………. (𝑖𝑣) From equation (𝑖) and (𝑖𝑣) we can say that emf and current are in phase. The phasor diagram is shown in fig 3. The variation of emf and current with time is shown in fig. 2. 𝐼 𝜀 Fig: Phasor diagram R 𝜀 = 𝜀0 sin 𝜔𝑡 𝐼 𝐼 𝑜𝑟 𝜀 𝜀 = 𝜀0 sin 𝜔𝑡 𝐼 = 𝐼0 sin 𝜔𝑡 𝜔𝑡 Fig: Variation of 𝐼 𝑜𝑟 𝜀 with 𝜔𝑡
- 9. A.C. through an inductor Suppose an inductor of inductance L is connected in series with 𝑎. 𝑐. source as shown in figure. 𝐼 is the current supplied by 𝑎. 𝑐. source. The instantaneous value of emf is given by 𝜀 = 𝜀0 sin 𝜔𝑡………. (𝑖) The induced emf across the inductor 𝜀𝐿 = −𝐿 𝑑𝐼 𝑑𝑡 Now from Kirchhoff’s 2nd law, 𝜀 + 𝜀𝐿 = 0 𝑜𝑟, 𝜀 − 𝐿 𝑑𝐼 𝑑𝑡 = 0 𝑜𝑟, 𝐿 𝑑𝐼 𝑑𝑡 = 𝜀 = 𝜀0 sin 𝜔𝑡
- 10. 𝑜𝑟, 𝑑𝐼 𝑑𝑡 = 𝜀0 sin 𝜔𝑡 𝐿 𝑜𝑟, 𝑑𝐼 = 𝜀0 𝐿 sin 𝜔𝑡 𝑑𝑡 The total current is obtained by integrating the above relation 𝐼 = 𝜀0 𝐿 sin 𝜔𝑡 𝑑𝑡 = 𝜀0 𝐿 sin 𝜔𝑡 𝑑𝑡 = 𝜀0 𝐿 − cos 𝜔𝑡 𝜔 = 𝜀0 𝜔𝐿 − cos 𝜔𝑡 = 𝜀0 𝜔𝐿 sin 𝜔𝑡 − 𝜋 2 ………. (ii) For I to be maximum,sin 𝜔𝑡 − 𝜋 2 = 1 𝐼0 = 𝜀0 𝜔𝐿 ………….. (iii) Therefore equation (ii) becomes, 𝐼 = 𝐼0 sin 𝜔𝑡 − 𝜋 2 ………. (iv)
- 11. In equation (iii), 𝜔𝐿 = 𝑋𝐿 acts as the resistance in 𝑎. 𝑐. circuit. It is known as inductive reactance. It’s unit and dimension is similar to that of resistance. From equation (𝑖) and (𝑖𝑣), we can say that emf leads the current by 𝜋 2 or current lag’s behind the emf by 𝜋 2 . The phasor diagram is shown in fig (c). The variation curve is shown in fig. (b). 𝜋 2 Fig: Phasor diagram 𝜀 𝑜𝑟 𝐼 Fig: Variation of 𝜀 𝑜𝑟 𝐼 with 𝜔𝑡 𝜀 = 𝜀0 sin 𝜔𝑡 𝐼 = 𝐼0 sin 𝜔𝑡 − 𝜋 2 O 𝜋 2 𝜋 2𝜋 3𝜋 2
- 12. Suppose a capacitor of capacitance C is connected in series with 𝑎. 𝑐. source as shown in figure. 𝐼 is the current supplied by 𝑎. 𝑐. source. The instantaneous value of emf is given by 𝜀 = 𝜀0 sin 𝜔𝑡………. (i) If q is the charge present to the capacitor at any instant of time t, then the potential difference between the plates of capacitor ε = 𝑞 𝐶 A.C. through a Capacitor:
- 13. 𝑞 = 𝑐𝜀 = 𝑐𝜀0 sin 𝜔𝑡 Differentiating this equation with respect to time, 𝑑𝑞 𝑑𝑡 = 𝑑 𝑐𝜀0 sin 𝜔𝑡 𝑑𝑡 𝑜𝑟, 𝐼 = 𝑐𝜀0 𝑑 sin 𝜔𝑡 𝑑𝑡 𝑜𝑟, 𝐼 = 𝑐𝜀0 cos 𝜔𝑡 𝜔 = 𝜔𝑐𝜀0 cos 𝜔𝑡 = 𝜀0 1 𝜔𝑐 cos 𝜔𝑡 = 𝜀0 1 𝜔𝑐 sin 𝜔𝑡 + 𝜋 2 …………. (ii) For I to be maximum, sin 𝜔𝑡 + 𝜋 2 = 1 𝐼0 = 𝜀0 1 𝜔𝑐 ………….. (iii) Therefore equation (ii) becomes, 𝐼 = 𝐼0 sin 𝜔𝑡 + 𝜋 2 ………. (iv)
- 14. In equation (iii), 1 𝜔𝑐 = 𝑋𝐶 acts as the resistance in 𝑎. 𝑐. circuit. It is known as capacitive reactance. It’s unit and dimension is similar to that of resistance. From equation (𝑖) and (𝑖𝑣), we can say that current leads the emf by 𝜋 2 or emf lag’s behind the current by 𝜋 2 . The phasor diagram is shown in fig (c). The variation curve is shown in fig. (b). Fig: Phasor diagram 𝜋 2 𝐼 𝜀 𝜀 𝑜𝑟 𝐼 Fig: Variation of 𝜀 𝑜𝑟 𝐼 with 𝜔𝑡 𝜀 = 𝜀0 sin 𝜔𝑡 𝐼 = 𝐼0 sin 𝜔𝑡 + 𝜋 2 O 𝜋 2 𝜋 2𝜋 3𝜋 2
- 15. Reactance: Capacitor and inductor are also acts as resistor in 𝑎. 𝑐. circuit. So the resistance offered by capacitor or inductor is known as reactance. It is also known as impure resistance. Its role in 𝑎. 𝑐. circuit is similar to resistor in 𝑑. 𝑐.. Its unit and dimension is similar to that of resistor. Note: For low frequency ie, 𝑓 = 0𝐻𝑧 For high frequency ie, 𝑓 = ∝ 𝑋𝐶 = 1 𝜔𝐶 = 1 2𝜋𝑓𝐶 = 1 0 =∝ 𝑋𝐶 = 1 𝜔𝐶 = 1 2𝜋𝑓𝐶 = 1 ∝ = 0 Hence, for low frequency or for 𝑑. 𝑐. capacitor acts as open path and for high frequency it acts as short circuit. Note: For low frequency ie, 𝑓 = 0𝐻𝑧 For high frequency ie, 𝑓 = ∝ 𝑋𝐿 = 𝜔𝐿 = 2𝜋𝑓𝐿 = 0 𝑋𝐿 = 𝜔𝐿 = 2𝜋𝑓𝐿 =∝ Hence, for low frequency inductor acts as short circuit and for high frequency it acts as open path.
- 16. A.C. through RL circuit Suppose a resistor of resistance R and inductor of inductance L are connected in series with 𝑎. 𝑐. source as shown in figure. The instantaneous value of current is given by 𝐼 = 𝐼0 sin 𝜔𝑡………. (𝑖) Due to flow of this current the induced emf is given by 𝜀𝐿 = −𝐿 𝑑𝐼 𝑑𝑡 Now using the Kirchhoff’s 2nd law in the fig. 𝜀 + 𝜀𝐿 = 𝐼𝑅 𝑜𝑟, 𝜀 − 𝐿 𝑑𝐼 𝑑𝑡 = 𝐼𝑅 𝑜𝑟, 𝜀 = 𝐼𝑅 + 𝐿 𝑑𝐼 𝑑𝑡 = 𝐼0 sin 𝜔𝑡 R + L 𝑑 𝐼0 sin 𝜔𝑡 𝑑𝑡
- 17. 𝑜𝑟, 𝜀 = 𝐼0 sin 𝜔𝑡 R + 𝐿𝜔 cos 𝜔𝑡 = 𝐼0 sin 𝜔𝑡 × R + cos 𝜔𝑡 × 𝑋𝐿 𝜔𝐿 = 𝑋𝐿 = 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 Let 𝑍 = 𝑅2 + 𝑋𝐿 2 𝑜𝑟, ε = 𝐼0Z sin 𝜔𝑡 𝑅 𝑍 + cos 𝜔𝑡 𝑋𝐿 𝑍 In the adjoining figure 𝑅 𝑍 = cos 𝜃 and 𝑋𝐿 𝑍 = sin 𝜃 ∴ 𝜀 = 𝐼0Z sin 𝜔𝑡 cos 𝜃 + cos 𝜔𝑡 sin 𝜃 = 𝐼0Z sin 𝜔𝑡 + 𝜃 …………. (ii) For 𝜀 to be maximum,sin 𝜔𝑡 + 𝜃 = 1 𝜀0 = 𝐼0Z………….. (iii) Therefore equation (ii) becomes, 𝜀 = 𝜀0 sin 𝜔𝑡 + 𝜃 ………. (iv) R 𝑋𝐿 𝜃
- 18. In equation (iii) 𝑍 = 𝑅2 + 𝑋𝐿 2 acts as resistance in 𝑎. 𝑐. circuit. It is known as impendence in RL- circuit. From equation (i) and (iv) we can say that emf leads the current by 𝜃 or current lags behind with emf by 𝜃 = tan−1 𝑋𝐿 𝑅 . The phasor diagram and variation curve is shown in fig. 𝜃 𝐼 𝜀 Fig: Phasor diagram
- 19. A.C. through RC circuit Suppose a resistor of resistance R and capacitor of capacitance C are connected in series with 𝑎. 𝑐. source as shown in figure. The instantaneous value of current is given by 𝐼 = 𝐼0 sin 𝜔𝑡………. (i) Due to flow of current the potential difference across resistor and capacitor are 𝑉𝑅 𝑎𝑛𝑑 𝑉𝐶 respectively. So from figure 𝜀 = 𝑉𝑅 + 𝑉𝐶 𝑜𝑟, 𝜀 = 𝐼𝑅 + 𝑞 𝐶 = 𝐼0 sin 𝜔𝑡 𝑅 + 1 𝐶 𝐼𝑑𝑡 = 𝐼0 sin 𝜔𝑡 𝑅 + 1 𝐶 𝐼0 sin 𝜔𝑡 𝑑𝑡 = 𝐼0 sin 𝜔𝑡 𝑅 + 1 𝐶 sin 𝜔𝑡 𝑑𝑡
- 20. = 𝐼0 sin 𝜔𝑡 𝑅 − 1 𝜔𝐶 cos 𝜔 𝑡 = 𝐼0 sin 𝜔𝑡 𝑅 − 𝑋𝐶 cos 𝜔 𝑡 ∵ 1 𝜔𝐶 = 𝑋𝐶 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 Let 𝑍 = 𝑅2 + 𝑋𝑐 2 ∴ 𝜀 = 𝐼0Z sin 𝜔𝑡 𝑅 𝑍 − cos 𝜔𝑡 𝑋𝐶 𝑍 In the adjoining figure 𝑅 𝑍 = cos 𝜃 and 𝑋𝐶 𝑍 = sin 𝜃 ∴ 𝜀 = 𝐼0Z sin 𝜔𝑡 cos 𝜃 − cos 𝜔𝑡 sin 𝜃 = 𝐼0Z sin 𝜔𝑡 − 𝜃 …………. (ii) For 𝜀 to be maximum,sin 𝜔𝑡 − 𝜃 = 1 𝜀0 = 𝐼0Z………….. (iii) Therefore equation (ii) becomes, 𝜀 = 𝜀0 sin 𝜔𝑡 − 𝜃 ………. (iv) R 𝑋𝑐 𝜃
- 21. In equation (iii) 𝑍 = 𝑅2 + 𝑋𝐶 2 acts as resistance in 𝑎. 𝑐. circuit. It is known as impendence in RC- circuit. From equation (𝑖) and (𝑖𝑣) we can say that current leads the emf by 𝜃 or emf lags behind with current by 𝜃 = tan−1 𝑋𝐶 𝑅 . The phasor diagram and variation curve is shown in fig. 𝜃 𝐼 𝜀 Fig: Phasor diagram
- 22. A.C. through LCR series circuit Suppose a resistor of resistance R and capacitor of capacitance C and inductor of inductance L are connected in series with 𝑎. 𝑐. source as shown in figure. The instantaneous value of current is given by 𝐼 = 𝐼0 sin 𝜔𝑡………. (i) Due to flow of current the potential difference across resistor and capacitor and inductor L are 𝑉𝑅 ; 𝑉𝐶 𝑎𝑛𝑑 𝜀𝐿 respectivey. So from figure 𝜀 + 𝜀𝐿 = 𝑉𝑅 + 𝑉𝐶 𝑜𝑟, 𝜀 − 𝐿 𝑑𝐼 𝑑𝑡 = 𝐼𝑅 + 𝑞 𝐶 or, ε = 𝐼0 sin 𝜔𝑡 𝑅 + 1 𝐶 𝐼𝑑𝑡 + 𝐿 𝑑𝐼 𝑑𝑡 = 𝐼0 sin 𝜔𝑡 𝑅 + 1 𝐶 𝐼0 sin 𝜔𝑡 𝑑𝑡 + 𝐿 𝑑 𝐼0 sin 𝜔𝑡 𝑑𝑡 = 𝐼0 sin 𝜔𝑡 𝑅 + 𝐿 𝑑 sin 𝜔𝑡 𝑑𝑡 + 1 𝐶 sin 𝜔𝑡 𝑑𝑡
- 23. 𝑜𝑟, 𝜀 = 𝐼0 sin 𝜔𝑡 R + 𝐿𝜔 cos 𝜔𝑡 − 1 𝜔𝑐 cos 𝜔𝑡 = 𝐼0 sin 𝜔𝑡 R + cos 𝜔𝑡 𝑋𝐿 − 𝑋𝑐 Let 𝑍 = 𝑅2 + 𝑋𝐿 − 𝑋𝐶 2 𝑜𝑟, ε = 𝐼0Z sin 𝜔𝑡 × 𝑅 𝑍 + cos 𝜔𝑡 × 𝑋𝐿 − 𝑋𝑐 𝑍 In the adjoining figure 𝑅 𝑍 = cos 𝜃 and 𝑋𝐿−𝑋𝐶 𝑍 = sin 𝜃 ∴ 𝜀 = 𝐼0Z sin 𝜔𝑡 cos 𝜃 + cos 𝜔𝑡 sin 𝜃 = 𝐼0Z sin 𝜔𝑡 + 𝜃 …………. (ii) For 𝜀 to be maximum,sin 𝜔𝑡 + 𝜃 = 1 𝜀0 = 𝐼0Z………….. (iii) Therefore equation (ii) becomes, 𝜀 = 𝜀0 sin 𝜔𝑡 + 𝜃 ………. (iv) R 𝑋𝐿 − 𝑋𝑐 𝜃
- 24. In equation (iii) 𝑍 = 𝑅2 + 𝑋𝐿 − 𝑋𝐶 2 acts as resistance in 𝑎. 𝑐. circuit. It is known as impendence in LCR- circuit. From equation (i) and (iv) we can say that emf leads the current by 𝜃 or current lags behind with emf by 𝜃 = tan−1 𝑋𝐿−𝑋𝐶 𝑅 . The phasor diagram and variation curve is shown in fig. 𝜃 𝐼 𝜀 Fig: Phasor diagram
- 25. Impedance: The total resistance offered by L; C ; R components in 𝑎. 𝑐. circuit is called impedance. It consists pure resistance R and impure resistance 𝑋𝐿 𝑎𝑛𝑑 𝑋𝐶 Known as reactance. It is denoted by Z. Its unit and dimension is similar to that of resistance. Its role in 𝑎. 𝑐. circuit is similar to resistor in 𝑑. 𝑐. circuit. Impedance in LCR series circuit is 𝑍 = 𝑅2 + 𝑋𝐿 − 𝑋𝐶 2 If the coil has material resistance r then 𝑍 = 𝑅 + 𝑟 2 + 𝑋𝐿 − 𝑋𝐶 2 Admittance: The reciprocal of impedance in 𝑎. 𝑐. circuit is called admittance. It is denoted by Y. 𝑖𝑒, 𝑌 = 1 𝑍 . Its unit is mho or siemens.
- 26. Resonance frequency The electric current in LCR series circuit is given by 𝐼 = 𝜀 𝑍 = 𝜀 𝑅2+ 𝑋𝐿−𝑋𝐶 2 For I to be maximum, the denominator of above expression should be minimum it is possible only when 𝑋𝐿 = 𝑋𝐶 𝑜𝑟, 𝜔𝐿 = 1 𝜔𝐶 𝑜𝑟, 𝜔2 = 1 𝐿𝐶 𝑜𝑟, 4𝜋2𝑓2 = 1 𝐿𝐶 𝑜𝑟, 𝑓2 = 1 4𝜋2𝐿𝐶 ∴ 𝑓 = 1 2𝜋 𝐿𝐶 ; This is an expression of resonance frequency.
- 27. Quality factor (Q-factor) The Q-factor of series resonant circuit is defined as the ratio of the voltage developed across the inductor or capacitor as resonance to the applied voltage or voltage across R. 𝑄 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝐿 𝑜𝑟 𝐶 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑅 = 𝐼𝑋𝐿 𝐼𝑅 𝑜𝑟, 𝐼𝑋𝐶 𝐼𝑅 = 𝜔𝐿 𝑅 𝑜𝑟, 1 𝜔𝑐𝑅 = 1 𝐿𝐶 𝐿 𝑅 𝑜𝑟, 1 1 𝐿𝐶 𝑐𝑅 ∵ 𝜔 = 2𝜋𝑓 = 2𝜋 1 2𝜋 𝐿𝐶 = 1 𝐿𝐶 = 1 𝑅 𝐿 𝐶 𝑜𝑟, 1 𝑅 𝐿 𝐶
- 28. Power consumed in a series LCR circuit The instantaneous value of emf is given by 𝜀 = 𝜀0 sin 𝜔𝑡………. (𝑖) The instantaneous value of current is 𝐼 = 𝐼0 sin 𝜔𝑡 ± 𝜙 ; where 𝜙 is phase difference between 𝜀 𝑎𝑛𝑑 𝐼. Therefore the instantaneous power 𝑃𝑖𝑛𝑠 = 𝜀𝐼 = 𝜀0 sin 𝜔𝑡 × 𝐼0 sin 𝜔𝑡 ± 𝜙 = 𝜀0𝐼0 sin 𝜔𝑡 sin 𝜔𝑡 ± 𝜙 The small amount of work done 𝑑𝑤 in small time dt is given by 𝑑𝑤 = 𝑃𝑖𝑛𝑠𝑑𝑡 = 𝜀0𝐼0 sin 𝜔𝑡 sin 𝜔𝑡 ± 𝜙 𝑑𝑡 = 𝜀0𝐼0 sin 𝜔𝑡 sin 𝜔𝑡 cos 𝜙 ± cos 𝜔𝑡 sin 𝜙 = 𝜀0𝐼0 𝑠𝑖𝑛2𝜔𝑡 cos 𝜙 ± sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙
- 29. The total work done in complete cycle of 𝑎. 𝑐. is obtained by integrating the above expression from 0 to T. 𝑊 = 0 𝑇 𝜀0𝐼0 𝑠𝑖𝑛2𝜔𝑡 cos 𝜙 ± sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙 𝑑𝑡 = 𝜀0𝐼0 cos 𝜙 0 𝑇 𝑠𝑖𝑛2𝜔𝑡 𝑑𝑡 ± sin 𝜙 0 𝑇 sin 𝜔𝑡 cos 𝜔𝑡 𝑑𝑡 = 𝜀0𝐼0 cos 𝜙 0 𝑇 1−cos 2𝜔𝑡 2 𝑑𝑡 ± sin 𝜙 0 𝑇 2sin 𝜔𝑡 cos 𝜔𝑡 2 𝑑𝑡 = 𝜀0𝐼0 2 cos 𝜙 0 𝑇 𝑑𝑡 − 0 𝑇 cos 2𝜔𝑡 𝑑𝑡 ± sin 𝜙 0 𝑇 sin 2𝜔𝑡 𝑑𝑡 = 𝜀0𝐼0 2 cos 𝜙 0 𝑇 𝑑𝑡 = 𝜀0𝐼0 2 cos 𝜙 𝑇 𝑜 𝑇 = 𝜀0𝐼0 2 × cos 𝜙 × 𝑇 − 0 = 𝜀0𝐼0 2 × cos 𝜙 × 𝑇
- 30. The average power consumed in complete cycle of 𝑎. 𝑐. is the ratio of total work done and time period. 𝑖𝑒 𝑃 = 𝑊 𝑇 = 𝜀0𝐼0 2𝑇 × cos 𝜙 × 𝑇 = 𝜀0𝐼0 2 × cos 𝜙 = 𝜀0 2 𝐼0 2 cos 𝜙 = 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙 The factor 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 is called virtual power or apparent power and the factor cos 𝜙 is called power factor. So the true power is the product of apparent power and power factor. The power factor cos 𝜙 is always positive and not more than 1. Power factor is the ratio of true power and apparent power.
- 31. Wattless and Wattful current The average power consumed in LCR series circuit is given by 𝑃𝑎𝑣 = 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙 In pure inductor or pure capacitor, the phase angle 𝜙 = 90° 𝑃𝑎𝑣= 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 90° = 0 Hence the current through pure L or pure C is called wattless current because it does not consumes any power. Again, in LCR series circuit at resonance 𝑖𝑒, 𝑋𝐿 = 𝑋𝐶 or, in pure resistor, the phase angle 𝜙 = 0° 𝑃𝑎𝑣= 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 0° = 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 Hence, purely resistor circuit consumes power. So, current flowing through R is called watt full current.