Question

Three Resistors in Parallel Three resistors are connected in parallel as shown in figure (a). A potential difference of 12.0V is maintained between points a and b. Three resistors connected in parallel. The Another circuit with three resistors and a voltage across each resistor is 12.0 V. battery. Is it equivalent to the circuit in (a)? a Ž 3.00 6.00 9.00 3.00 6.00 09.00 (a) Calculate the equivalent resistance of the circuit. SOLUTION Conceptualize Figure (a) shows that we are dealing with a simple parallel combination of three resistors. Notice that the current I splits into three currents 11, 12, and Iz in the three resistors. Categorize This problem can be solved with rules developed in this section, so we categorize it as a substitution problem. Because the three resistors are connected in parallel, we can use the rule for resistors in parallel to evaluate the equivalent resistance. Analyze Use the equation for resistors in parallel to find Reg (in): Ω Req = 1.64 (b) Find the current (in A) in each resistor. SOLUTION The potential difference across each resistor is 12.0 V. Apply the relationship AV = IR to find the currents: 11 A 12 AV R1 AV R₂ AV R3 12.0 V = 4 3.00 12.0 V = 2 6.002 12.0 V 1.33 9.00 2 А 13 - А

Answer 1

Solution : For the exercise problem given below first we need to obtain the value of unknown resistor R using the given data provided in the question statement then using basic form of Ohm's law the current through each parallel connected resistor must be same such that the power consumed by each resistor or delivered by the source element to each resistor is calculated as shown below-

Solution for the exercise problem: As the three equal resistors each are having the value of 'R' ohm are connected in parallel across the 18 Volt battery as shown in the figure- I= 1.90 A a R 18V b

Given that the current supplied by the battery- I = 1.90 Ampere Now, the equivalent resistance seen from the battery terminals is given as- f + I Req -14 I = 1.9A R R lov • Rea Dea ناام 2 a Req R 2 3 As equivalent resistance (Req) seen from the battery terminals is equal to the ratio of voltage impressed across the given network to the current supplied to the given network is given as- Req 3 Л. V ก where, V= voltage supplied by the battery. I = Current delievered by the battery.

From equation (2) and (3) we get- هامو) als R 3x18 1.9 54 R = 540 r 19 therefore, each resistor in the circuit is equal to R = 540 ohm 19 a JER R JOV ISR 12 J b Now, voltage across each resistor, is given as- AV Va - Vo = 18 volt As, all the resistors are equal therefore, the current supplied to each parallel connected resistor is same. Now the currents is given as - Δν J = 1 - 1 Ampere toxig 540 30 - (540/19) Jy = Io = Ig 19 Ampere 30

therefore, power delivered to each resistor in the circuit so Cove or (1, R = 1, R = I²R) Watt R. PR = IR PR 19 חן 2 х 540 19 Watt Po 11.4 watt hence power delivered to each resistor in the circuit is equal to 11.4 Watt Answer

Solution (a,b and c): For the circuit shown in figure- 3.00 2 6.00 9.00 a) Calculation of equivalent resistances- As equivalent resistance (Req) seen from the battery terminals is equal to the ratio of voltage impressed across the given network to the current supplied to the given network is given as- As voltage across each resistor is equal to - Af AV = Va-VD Au=12volt t Req v r I + I 62 + 1 ar ) 32 = 1.6362 Req u 1.642

OLNOSS (b) now as all the resistor connected in the circuit is in parallel connection therefore vachango each element is equal - It is equal to - 12 - 4 Ampere R RI I = 4 Ampere NOW I = Vab AV equal to to Iga Vab R2 AV - 12 6 = 2 Ampere I = 2 Ampere R2 I is equal to Iz = Vab SAV Ro 12 9 1:33 Ampere. Iz = 1:33 Ampere R3 (C) Poner delievered to each resistor -P3r = IRR wart (4) x3 P, = 48 watt Pq = Por = 12 R2 Watt = ( 27 x 6 → P₂ = Pgr = 7% Rg Watt (1.33 x 9 123 =l6 watt P2 = 24 watt Now, Total Power deliévered to the resisters – P, XP 2+P2 Watt Protal 40+24 +16 = 88 watt P