Question

determine current through each resistor ​

Answer 1

Answer:

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Explanation:

the current flows through open and closed circuit the current flows of 14 volt

Answer 2

These type of problems can be solved in many-many ways. One way is below:

Follow Kirchhoff ’s Current &Voltage Law along with Ohms Law.

Break the circuit first i.e.

Putting it this way will make it easy, illustration wise. Start calculating series and parallel registers now. J1 and J2 are your two Junctions.

Resisters R2, R3, R4 are in series. Therefore their total Resistance is 20Ω. Now consider this 20Ω register as R7. This R7 is parallel with R6.

Using reciprocal of reciprocal (formula):

We get the value of both parallel resistors (R7 & R6) as 8.57Ω. We will name it as R8. So now we have:

Total resistance (load) of the circuit is 5+8.57+5 = 18.57Ω (resistors in series will addup)

Total Voltage = 15 V

Total Resistance = 18.57 Ω.

Therefore (Ohm’s Law) I (current) = V / R = 15 / 18.57 = 0.81 Amperes.

If you notice, we have three segments here. One is from 15v supply to J1 (illustration wise), second is from J1 to J2 and third is from J2 to Grnd. Using current above, calculate the voltages for these three segments. This would be

To check, add the three voltages i.e. 4.05 + 6.94 + 4.05 = 15 volts (rounded)

Now we will build back the circuit, in the reverse order.

The individual segments’ currents are calculated using Ohm’s law.

For R7 I = V / R = 6.94 / 20 = 0.347A and R6 is 0.462A

We will concentrate on the segment J1 - J2.

According to Kirchoff’s Current & Voltage law - putting it simply - in Parallel circuits, the currents divide and in Series circuits the voltage divide. Therefore we have used 6.94 v (the voltage of that segment) to calculate the divided current between R7 and R6 parallel resistors.

Building up or breaking up R7 / 20Ω resistor again we get our three resistors R2, R3, R4 back.

Now voltage for segment J1-J2 is 6.94 v. and the current flowing through R2, R3 and R4 resistors is 0.347A only (as calculated above). Now we have to calculate the individual voltages of the resistors in series i.e. R2, R3, R4. Accordingly, V=IR and…

Voltage for R2 = 0.347A x 5Ω = 1.735v

Voltage for R3 = 0.347A x 10Ω = 3.47 v and

Voltage for R4 = 0.347 x 5 Ω = 1.735 v

Adding the three voltages we will get 1.735+3.47+1.735 = 6.94 volts (the total segment voltage)

and Now we have

Total current of three segments will remain 0.81 amps (segment J1-J2 0.347 + 0.462 = 0.81 amps.