Shown in the figure at the right are two resistors connected in parallel with a

Question

Shown in the figure at the right are two resistors connected in parallel with a battery. The current through the top resistor is three times the current through the bottom resistor. If each resistor was connected by itself to the battery, how would the current through each compare to the other? Explain fully and give evidence. If the two resistors were connected in series with the battery, how would the currents through each compare to the other? Explain fully and give evidence. If the two resistors were connected in series with the battery, how would the current from the battery compare to the current from the battery in the circuit above? Explain fully and give evidence.

Explanation / Answer

Let the battery be of V volts, top Resistance R1 and bottom Resistance R2.

Given is I1 = 3*I2 => V/R1 = 3* V/R2 => R2 = 3*R1

1. If each resistor is connected by itself to the battery, then the current through R1 will be 3 times the current through R2 because with same battery, the resistance is one third for R1 as compared to R2

2. If the two resistors are connected in series then same amount of current will flow through both of them = V/(R1+R2)

3. Let R1 = R , R2 = 3*R1 = 3R

Current from battery in case of both in series with battery = V/(R+3R) = V/4R

For given diagram parallel connection

Current through battery = V*(R+3R)/(R*3R) = 4V/3R

The current in parallel connection is 16/3 times the current in series connection.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.