The battery in the figure below has negligible internal resistance 3 S2 6V 2 52

Question

The battery in the figure below has negligible internal resistance 3 S2 6V 2 52 4 S2 252 1) (a) Find the current in each resistor. A (3 resistor) Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question 2) A (4 resistor) Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question 3) A (vertical 2 Q resistor) Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question 4) A (diagonal 2 Q resistor) Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question 5) (b) Find the power delivered by the battery. W Submit

Explanation / Answer

4 ohms and 2 ohms and 2 ohms are in parallel.

if their equivalent resistance is R,

then 1/R=(1/2)+(1/2)+(1/4)

==>R=0.8 ohms

it is in series with 3 ohms.

hence total resistance=3+R=3.8 ohms

current in the circuit=voltage/total resistance

=6/3.8=1.5789 A

hence current through 3 ohms resistor is 1.5789 A

voltage drop across 3 ohms resistor=resistance*current

=3*1.5789=4.7368 volts

so voltage across the parallel combination=6-voltage across 3 ohms resistor

=6-4.7368

=1.2632 volts

so current through vertical 2 ohms=current through diagonal 2 ohms=voltage / resistance

=1.2632/2=0.63158 A

current through 4 ohms resistor=voltage/resistance

=1.2632/4=0.3158 A

power delivered by the source=voltage*total current

=6*1.5789=9.4737 W

so answers are:

part 1:

1.5789 A

part 2:

0.3158 A

part 3:

0.63158 A

part 4:

0.63158 A

part 5:

9.4737 W

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