circuit question
why is the answer C?
I have considered power and it is larger for 2 than 1 and larger for 4 than 3.
This is simply because of the one difference of series and parallel arrangement.
But why is the power for 4 and 1 the same?
For 1, power is ((2V)^2)/(2R)=(2)(V^2)/(R) taking V as EMF from one cell and R as resistance of one resistor.
That means power for 4 is the same.
So for 4, do I start from one of the cell, say the top one, and complete a loop passing by one of the resistors, say the top one, and do the same for the bottom cell and bottom resistor, and then add the powers for each loop?
Cheers
I have considered power and it is larger for 2 than 1 and larger for 4 than 3.
This is simply because of the one difference of series and parallel arrangement.
But why is the power for 4 and 1 the same?
For 1, power is ((2V)^2)/(2R)=(2)(V^2)/(R) taking V as EMF from one cell and R as resistance of one resistor.
That means power for 4 is the same.
So for 4, do I start from one of the cell, say the top one, and complete a loop passing by one of the resistors, say the top one, and do the same for the bottom cell and bottom resistor, and then add the powers for each loop?
Cheers
From P = V2/R and P = E/t we have (E × R)/V2 = t.
For case 1, we have RT = 2r (because in series) and VT = 2v (again, because in series, so just sum), where r and v are the resistance of one resistor and the potential difference across one cell respectively. Therefore,
(E × 2r)/((2v)2) = (Er)/(2v2) = t1.
For case 4, we have RT = ½r (because in parallel) and VT = v (as from Kirchhoff’s voltage law, the voltage from the positive end of the two cells is just v, from where ever you take the loop from just those two cells). Therefore,
(E × ½r)/v2 = (Er)/(2v2) = t4 (= t1).
For case 1, we have RT = 2r (because in series) and VT = 2v (again, because in series, so just sum), where r and v are the resistance of one resistor and the potential difference across one cell respectively. Therefore,
(E × 2r)/((2v)2) = (Er)/(2v2) = t1.
For case 4, we have RT = ½r (because in parallel) and VT = v (as from Kirchhoff’s voltage law, the voltage from the positive end of the two cells is just v, from where ever you take the loop from just those two cells). Therefore,
(E × ½r)/v2 = (Er)/(2v2) = t4 (= t1).
(edited 1 month ago)
Hello Harlem!
My paper is as follows:
Please note: In the text given below, "E" represents the total energy that each battery can supply before it is discharged. This energy is constant for all the batteries in the circuit. The time ( tx ) that the battery can supply current is proportional to the total energy (E) divided by the power ( Px ) consumed by the circuit.
Circuit 1: One cell with three resistors in series. Total resistance (R1 = 3R) Current (I1 = V/3R) Power (P1 = I1^2 x 3R = V^2/3R) Time (t1 proportionality to E/P1 = 3R x E/V^2)
Krgds,
Sandro
My paper is as follows:
Please note: In the text given below, "E" represents the total energy that each battery can supply before it is discharged. This energy is constant for all the batteries in the circuit. The time ( tx ) that the battery can supply current is proportional to the total energy (E) divided by the power ( Px ) consumed by the circuit.
Circuit 1: One cell with three resistors in series. Total resistance (R1 = 3R) Current (I1 = V/3R) Power (P1 = I1^2 x 3R = V^2/3R) Time (t1 proportionality to E/P1 = 3R x E/V^2)
Circuit 2: Two cells in parallel, each with one resistor. Total resistance (R2 = R/2) Current (I2 = 2V/R) Power (P2 = I2^2 *x R/2 = 4V^2/R) Time (t2 proportionality to E/P2 = R x E/4V^2)
Circuit 3: Two cells in series with two resistors in parallel. Total resistance (R3 = R/2) Current (I3 = 2V/R) Power (P3 = I3^2 * R/2 = 4V^2/R) Time (t3 proportionality to E/P3 = R x E/4V^2)
Circuit 4: Three cells in parallel, each with one resistor. Total resistance (R4 = R/3) Current (I4 = 3V/R) Power (P4 = I4^2 x R/3 = 9V^2/R) Time (t4 proportionality to E/P4 = R x E/9V^2)
Comparison, you must analyse the E/Pcircuit = R x E/NnV^2 for the four circuits:
t1 = t4;
t2 = t3;
t1 = t4 < t2 = t3
So, the option "C" is correct: t1 = t4 < t2 = t3.
Circuit 3: Two cells in series with two resistors in parallel. Total resistance (R3 = R/2) Current (I3 = 2V/R) Power (P3 = I3^2 * R/2 = 4V^2/R) Time (t3 proportionality to E/P3 = R x E/4V^2)
Circuit 4: Three cells in parallel, each with one resistor. Total resistance (R4 = R/3) Current (I4 = 3V/R) Power (P4 = I4^2 x R/3 = 9V^2/R) Time (t4 proportionality to E/P4 = R x E/9V^2)
Comparison, you must analyse the E/Pcircuit = R x E/NnV^2 for the four circuits:
t1 = t4;
t2 = t3;
t1 = t4 < t2 = t3
So, the option "C" is correct: t1 = t4 < t2 = t3.
Krgds,
Sandro
(edited 1 month ago)
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