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In the circuit shown in Fig. E26.18,ε=36.V,R1=4.0Ω,R2=6.0Ω,R3=3.0Ω(a) What is the potential difference Vab between points a and b when the switch S is open and when S is closed? (b) For each resistor, calculate the current through the resistor with S open and with S closed. For each resistor, does the current increase or decrease when S is closed?

Short Answer

Expert verified

(a) The potential difference between points a and b when S is open is 24 V.

(b) The potential difference between points a and b when S is closed is 16 V.

When S is open I1=0Aand I2=I3=4Aand when S is closed I1=4Aincreases I2=2.67AandI3=6.67Aincreases

Step by step solution

01

Step:1 Concept

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

02

Step 2:Determine the potential difference between points a and b when S is open

We are givenR1=4.0Ω,R2=6.0Ω,R3=3.0Ω

The emf of the battery is E=36. V

(a)

We Want V, when S is open and when it is closed.

When S is open, the current flow only in R and R, and the voltage drop of the battery is due to R2 and R3 As both these resistors are in series, therefore, they have the same current and We could get the current by using Ohm‘s law in the next form.

role="math" localid="1655716159141" V=I(R2+R3)I=VR2+R3..........................(1)

Now let us plug our values for V, R2, and R3 into equation (1) to get I which will help us to find the required voltage,

role="math" localid="1655716064912" I=36V6Ω+3Ω=4A

The voltage across ab is due to the voltage of R2 so this voltage could be calculated by using Ohm's law in the next form,

role="math" localid="1655715865929" Vab=IR2=(4.0A)×(6.0Ω)=24.0V

When S has closed the current flow in all resistors where R1 and R2 are in parallel and their combination could be

Therefore the potential difference between points a and b when S is open is 24.0 V

03

Determine the  potential difference between points a and b when S is closed 

Calculated the equivalent resistance when R1 and R2 are in parallel combination,

R12=R1R2R1+R2=4Ω×6Ω4Ω+6Ω=2.4Ω

The current in the circuit will change, and the new current (I2=I3)could be calculated using equation (1) in the next form

I12=36V2.4Ω+3Ω=6.67A

As R1 and R2 are in parallel, therefore, the voltage across ab represents the voltage across their combination and it could be calculated by Ohm's law in the next form,

Vab=I12R12Vab=(6.67A)×(2.4Ω)Vab=16V

Therefore the potential difference between points a and b when S is closed is 16 V

04

Determine when S is the open effect on current

(b)

Only in R2 and R3 and as both resistors are in series, therefore, the current is the same for both of them Where from part (a) the current is I2=I3=4A

When S is closed, the current flows in R3 are the same as in part (a)

I1=0A

But for the combination Rm the current is cut into both resistors.

Where We can use the voltage across ab for each resistor.

To obtain the current in R1 and R2 using Ohm's law

I1=VabR1I1=16V4ΩI1=4AI2=VabR2I2=16V6ΩI2=2.67A

Therefore when S is open I1=0Aand I2=I3=4Aand when S is closed I1=4Aincreases I2=2.67AandI3=6.67A increases

The current through R1 and R3 increases while through R2 it decreases.

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