Four lamps of 200 w, 100 w, 60 w and 40 w are connected in parallel to a power supply of 220 v. calculate :- (i) total current consumed (ii) total resistance of this arrangement (iii) the cost of keeping them lighted for 7 hours daily for 30 days, the cost of electricity being 30 paise per unit.
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Answer:
3600/-
Explanation:
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Answer with Explanation:
Let [tex]P_1=200W[/tex]
[tex]P_2=100 W[/tex]
[tex]P_3=60 W[/tex]
[tex]P_4=40 W[/tex]
Total power=P=[tex]P_1+P_2+P_3+P_4=200+100+60+40=400 W[/tex]
1 KW=[tex]1000 W[/tex]
P=[tex]\frac{400}{1000}=0.4 KW[/tex]
Potential difference=V=220 V
We know that
Power, P=VI
(i).In parallel combination , potential difference remains same but current are different passing through different device.
[tex]P_1=VI_1[/tex]
[tex]I_1=\frac{200}{220}=0.91 A[/tex]
[tex]I_2=\frac{P_2}{V}=\frac{100}{220}=0.455 A[/tex]
[tex]I_3=\frac{P_3}{V}=\frac{60}{220}=0.27 A[/tex]
[tex]I_4=\frac{40}{220}=0.182 A[/tex]
(ii)Total current in the circuit= [tex]I=I_1+I_2+I_3+I_4=0.91+0.455+0.27+0.182\approx 1.82 A[/tex]
We know that
[tex]R=\frac{V}{I}[/tex]
Using the formula
[tex]R=\frac{220}{1.82}=120.9\Omega[/tex]
(iii) Time, t=[tex]7\times 30=210 hours[/tex]
E=Pt
Using the formula
[tex]E=0.4\times 210=84 KWh[/tex]
1 KWh= 1unit
Cost of 1 unit=30 paise=[tex] Rs\frac{30}{100}[/tex]
Rs 1=100 paise
Cost of 84 unit=[tex]84\times \frac{30}{100}=Rs 25.2[/tex]
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