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A resistance and a capacitor connected in series across a 250V supply draws 5A at 50Hz. When frequency is increased to 60Hz, it draws 5.8A. Find the values of R & C. Also find active power and power factor in both cases.

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#### Solution

𝐼_{1}=5𝐴 V = 250V 𝑓_{1}=50𝐻𝑧 𝑓_{2}=60𝐻𝑧 𝐼_{2}=5.8𝐴

(1) Values of R and C.

For 𝑓_{1}=50𝐻𝑧 𝑍_{1}=`V/I_1=250/5`= 50Ω

`Z_1^2=(R^2+(1/(2pif_1c))^2)=(R^2+(1/(100piC))^2)`

`R^2+(1/(100piC))^2=2500`…………….(1)

For 𝑓_{1}=60𝐻𝑧 `Z_2=V/I_2=250/5.8` = 43.1Ω

`Z_2^2=(R^2+1/(2pif_1c))=(R^2+(1/(100piC))^2)`

`R^2+(1/(120piC))^2`=1857.9Ω …………….(2)

From (1) and (2) we get,

R = 19.96Ω C = 69.4𝝁𝑭

(2) Power draw in the both cases.

𝑃_{1}= 𝐼_{1}^{2}𝑅=5^{2}×19.96 = 499w 𝑃_{2}= 𝐼_{2}^{2}𝑅=5.8^{2}×19.96 = 671.45w

𝑷_{𝟏}=𝟒𝟗𝟗𝒘 and 𝑷_{𝟐}= 671.45w

(3) Power factor in both cases:-

P_{1}=VI_{1}cosφ_{1}

499=250×5×cos𝜑_{1} => cos𝜑_{1}=0.3992

𝝋_{𝟏}=𝟔𝟔.𝟒𝟕𝟏°

P_{2}=VI_{2}cosφ_{2}

671.45=250×5.8×cos𝜑_{2} => cos𝜑_{2}=0.463

𝝋_{𝟏}=𝟔𝟐.𝟒𝟏𝟒𝟔°