Question

a. Circuit Element AV (V) (4) R(2) Battery 12 3.6 89 R1 -12 -0.71 17 R2 -12 -0.413 29 R3 -12 -0.279 43 b. What is the relationship among the potential drops across each resistor when the resistors are in

Answer 1

Answer:

A. Battery Voltage = 12 V

Voltage across parallel resistors is always same.

we see that all resistor have potential difference of -12 V and magnitude is equal to the Battery Voltage but in opposite polarity.

Since all resistor have unequal resistance but have equal voltage across themselves. This means all resistors are connected in parallel as shown below.

A 1: -1.40 A PR2 (V PR1 EV: -12.0 V V1 12V R1 1722 R2 29.22 R3 4322 + Ref1

D. since all three resistor are connected in parallel.

Their equivalent parallel resistance is given by,

Rnet 1 R1 RI 1 R2 + 1 R3

$R_{net} = \frac{1}{\frac{1}{17}+\frac{1}{29}+\frac{1}{43}}$

$R_{net} = 8.579 \Omega$   (answer)

TO VERIFY:

Add all current flowing through each resistor.

Itotal R1 + R2 + 1 R3

Voltage across each resistor in parallel, V = -12 V

So, total resistance as seen by Battery = $R_{net} =\frac{V}{ I_{total}} =\frac{-12}{-1.402} = 8.559 \Omega$

Total resistance obtained is approximately equal from both methods.

power delivered to the circuit = V * I total = -12 * -1.402 = 16.824 W

E. if fourth resistor is added in parallel with the three resistors, then Net resistance of the Circuit will decrease.

By ohm's law, battery voltage is constant and resistance has decreased so the total current flowing out of the battery will increase. $I = \frac{V}{R}$

V =constant voltage. if resistance increases, total current decreses and vice-versa.

Power delivered by the Battery will also increase.

Value of fourth resistor is not given.

Lets take R4 = 10 ohm

New equivalent parallel resistance is given by,

$R_{net} = \frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+ \frac{1}{R4}}$

$R_{net} = \frac{1}{\frac{1}{17}+\frac{1}{29}+\frac{1}{43}+\frac{1}{10}}$

$R_{net} =4.615 \Omega$

NET RESISTANCE OF CIRCUIT DECREASED.

NEW TOTAL CURRENT FLOWING FROM THE BATTERY INTO THE CIRCUIT IS,

$I_{total} = \frac{V}{R_{net}} = \frac{-12}{4.615}= -2.60 A$

New power delivered by the battery = V * I total = -12 * -2.60 = 31.2 W

1: -2.60 A V: -12.0 V PR1 + - Ref1 А PR2 R4 102 V1 12V R1 1722 R2 3290 R3 $430 V Ref1