Question

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.

Answer 1

Answer:

3600/-

Explanation:

lagta toh hai.....

pata Nahi

Answer 2

Answer with Explanation:

Let $P_1=200W$

$P_2=100 W$

$P_3=60 W$

$P_4=40 W$

Total power=P=$P_1+P_2+P_3+P_4=200+100+60+40=400 W$

1 KW=$1000 W$

P=$\frac{400}{1000}=0.4 KW$

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.

$P_1=VI_1$

$I_1=\frac{200}{220}=0.91 A$

$I_2=\frac{P_2}{V}=\frac{100}{220}=0.455 A$

$I_3=\frac{P_3}{V}=\frac{60}{220}=0.27 A$

$I_4=\frac{40}{220}=0.182 A$

(ii)Total current in the circuit= $I=I_1+I_2+I_3+I_4=0.91+0.455+0.27+0.182\approx 1.82 A$

We know that

$R=\frac{V}{I}$

Using the formula

$R=\frac{220}{1.82}=120.9\Omega$

(iii) Time, t=$7\times 30=210 hours$

E=Pt

Using the formula

$E=0.4\times 210=84 KWh$

1 KWh= 1unit

Cost of 1 unit=30 paise=$Rs\frac{30}{100}$

Rs 1=100 paise

Cost of 84 unit=$84\times \frac{30}{100}=Rs 25.2$

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