1. Problem 1: A small flashlight bulb draws 300mA from its 1.5V battery. (a) What is the resistance of the bulb? - Given: Current $I = 300mA = 0.3A$, Voltage $V = 1.5V$ - Formula: Ohm's Law $V = IR$ which rearranges to $R = \frac{V}{I}$ - Calculation: $$R = \frac{1.5}{0.3}$$ $$R = 5\ \Omega$$ (b) If the voltage dropped to 1.2V, how would the current change? - Given new voltage $V_{new} = 1.2V$, resistance $R = 5\ \Omega$ (assumed constant) - Using Ohm's Law again: $$I_{new} = \frac{V_{new}}{R} = \frac{1.2}{5}$$ $$I_{new} = 0.24A = 240mA$$ 2. Problem 2: Find the resistance of a circuit that draws 0.06 amperes with 12 volts applied. - Given: $I = 0.06A$, $V = 12V$ - Using Ohm's Law: $$R = \frac{V}{I} = \frac{12}{0.06}$$ $$R = 200\ \Omega$$ 3. Problem 3: If a small appliance is rated at a current of 10A and a voltage of 120V, find the power rating. - Given: $I = 10A$, $V = 120V$ - Formula for power: $P = VI$ - Calculation: $$P = 120 \times 10$$ $$P = 1200\ W$$ Final answers: 1(a) Resistance = $5\ \Omega$ 1(b) New current = $240mA$ 2 Resistance = $200\ \Omega$ 3 Power rating = $1200W$