in the circuit seen here, the resistor has a resistance of 3 ohms. if no change in the battery size occurs…

in the circuit seen here, the resistor has a resistance of 3 ohms. if no change in the battery size occurs, what will happen to the current if the resistance increases to 6 ohms?\na it will increase by a factor of 2.\nb it will decrease by a factor of 2.\nc it will increase by a factor of 6.\nd it will decrease by a factor of 3.

in the circuit seen here, the resistor has a resistance of 3 ohms. if no change in the battery size occurs, what will happen to the current if the resistance increases to 6 ohms?\na it will increase by a factor of 2.\nb it will decrease by a factor of 2.\nc it will increase by a factor of 6.\nd it will decrease by a factor of 3.

Answer

Answer:

B. It will decrease by a factor of 2.

Explanation:

Step1: Recall Ohm's Law

$V = IR$, where $V$ is voltage, $I$ is current and $R$ is resistance. Since the battery size doesn't change, $V$ is constant. So, $I=\frac{V}{R}$.

Step2: Calculate initial and final current

Let the initial resistance $R_1 = 3\Omega$ and final resistance $R_2=6\Omega$. Initial current $I_1=\frac{V}{R_1}$, final current $I_2=\frac{V}{R_2}$. Substitute the values: $I_1=\frac{V}{3}$ and $I_2=\frac{V}{6}$.

Step3: Find the ratio of currents

$\frac{I_2}{I_1}=\frac{\frac{V}{6}}{\frac{V}{3}}=\frac{V}{6}\times\frac{3}{V}=\frac{1}{2}$. So, $I_2=\frac{1}{2}I_1$, which means the current decreases by a factor of 2.