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A triangular array of resistors is shown in Fig. E. What current will this array draw from a 35.0-V battery having negligible internal resistance if we connect it across

(a) ab ;

(b) bc;

(c) ac ;

(d) If the battery has an internal resistance of 3.00Ω, what current will the array draw if the battery is connected across bc?

Short Answer

Expert verified

(a)Iab=3.5A

(b)Ibc=4.5A

(c)Iac=3.15A

(d)Ibc=3.25A

Step by step solution

01

Definition of equivalent resistance

The equivalent resistance is where the resistance connected either in parallel or series is calculated.

02

Determine the current through 

Let R1=15Ω,R2=10Ω, and R3=20Ω. The voltage of the battery is V =35.0 V.

a)

When the battery is connected across ab, the two resistorsR2and R3will be in series as shown in figure 1 and their combinationR23=R2+R3=30Ωis in parallel withR1.

Where the current through resistors connected in parallel is the sum of the currents through the individual resistorsR1andR23, so the current acrosswill be given by Ohm’s law.

It=Iab=VReq (1)

WhereReqrepresents the equivalent resistance in the circuit and it can be given as:

Req=R1R23R1+R23=(15)(30)15+30=10Ω

Plug these values into equation (1),

lab=VReq=35.0V10Ω=3.5A

The current flowing through ab is 3.5 A.

03

Determining current across bc

b)

As shown in figure 2, when the battery is connected across bc, the two resistors R1and R3will be in series and their combination R13=R1+R3=35Ωis in parallel with R2.

With the same steps as in part (a), the equivalent resistance can be given as:

Req=R2R13R2+R13=(10)(35)10+35=7.78Ω

Plug these values in equation (1),

Ibc=VReq=35.0V7.78Ω=4.5A

The current flowing through bc is 4.5 A.

04

Determining current across  

c)

With the same steps as above, the equivalent resistance is given as:

Req=R3R12R3+R12=(20)(25)20+25=11.11Ω

Plug these values in equation (1),

Iac=VReq=35.0V11.11Ω=3.15A

The current flowing through ac is 3.15 A.

05

Determining the current through  

d)

Here, r=3Ω

In part (b), the voltage drop is only due to Reqbut here it is due to Reqand r, so the current across bc is related to the emf of the battery, and its internal resistance is given as:

Ibc=εReq+r (2)

Where Req=7.78Ω, Now, plug these values in equation (2),

Ibc=35.0V7.78Ω+3Ω=3.25A

The current decreases due to the internal resistance of the battery.

Thus, the current through bc is 3.25 A.

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