Question

Exit ticket: A lamp is plugged into a 110 Volt electrical outlet. There is a 9 Watt LED bulb in the lamp. a. What is
the current in the bulb? b. What is the resistance of the bulb?

Answer 1

Answer:

(a) 0.081 A (b) 1358.02 ohms

Explanation:

Voltage, V = 110 volt

Power of a LED bulb, P = 9 Watt

(a) Let the current is I. The formula for ower in terms of voltage and power is given by :

P = VI

$I=\dfrac{P}{V}\\\\I=\dfrac{9}{110}\\\\=0.081\ A$

(b) Let R is the resistance of the bulb. Using Ohm's law as follows :

V = IR

$R=\dfrac{V}{I}\\\\R=\dfrac{110}{0.081}\\\\=1358.02\ \Omega$

Hence, this is the required solution.

Answer 2

(a) The current in the bulb is 82mA

(b) The resistance of the bulb is 1340 Ω

Given information:

Supply voltage, V = 110V

The power rating of the bulb, P = 9 W

Electrical Power:

(a) Current in the bulb:

The electrical power is given by:

P = VI

where V is the voltage, and I is the current

So, I = P/V

I = 9W / 110V = 0.082A

I = 82 mA

(b) Resistance of the bulb:

According to the Ohm's Law:

The relation between voltage (V), current (I), and resistance (R) is given by:

V = IR

110 = 0.082R

R = 1340 Ω

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