Question

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential difference across resistor 1? Express your answer with the appropriate units. V1 = 2.1 V Previous Answers Correct Part BPart complete If you decrease the value of R1 what happens to the current in the circuit? Current in the circuit decreases. Current in the circuit increases. Current in the circuit stays the same. Previous Answers Correct Part CPart complete If you decrease the value of R1 what happens to the potential difference across resistor 1? Potential difference across resistor 1 decreases. Potential difference across resistor 1 stays the same. Potential difference across resistor 1 increases.

Answer 1

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$.

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

Part A

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$. The voltage across the two resistor, combined, is equal to $\rm 6\; V$. Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$.

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$. Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$.

Part B

Since the equivalent resistance is equal to $R_1 + R_2$, when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$. When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

Part C

Keep in mind that if two resistors are connected in series,

$I(R_1) = I(\text{Circuit}) = I(R_2)$.

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

$V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V$,

when the voltage across R2 increases, the voltage across R1 will decrease.