An electric geyser is connected to a 220 ; V | vedclass

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Question

(C) Given: Voltage $(V)$ = $220 \; V$, Energy $(E)$ = $4.4 \; kWh = 4400 \; Wh$, Time $(t)$ = $1 \; hour$.First, calculate the power $(P)$ of the geys

Vedclass · Accepted Answer

$11 \; \Omega$

Vedclass · Answer

(C) Given: Voltage $(V)$ = $220 \; V$, Energy $(E)$ = $4.4 \; kWh = 4400 \; Wh$, Time $(t)$ = $1 \; hour$.First, calculate the power $(P)$ of the geyser using the formula $P = E / t$.$P = 4.4 \; kWh / 1 \; h = 4.4 \; kW = 4400 \; W$.Next, use the power formula $P = V^2 / R$ to find the resistance $(R)$.$R = V^2 / P$.$R = (220 \; V)^2 / 4400 \; W$.$R = (220 	imes 220) / 4400$.$R = 48400 / 4400 = 11 \; \Omega$.Therefore, the resistance of the geyser is $11 \; \Omega$.

An electric geyser is connected to a $220 \; V$ line and consumes $4.4 \; kWh$ of electrical energy in $1 \; hour$. What is the resistance of the geyser?

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