Question

.1.calculate the following

a.effective resistance of two 8ohm. resistors in the combination
b.current flowing through 4ohm resistors
c.potential differences across 4ohm resistors ​

Answer 1

Answer:

Given,

Resistor,R1 = 4ohm

Resistor,R2 = 8ohm

Resistor,R3 = 8ohm

Potential Difference,d = 8volts.

Refer to the attachment for diagram

i)Since we know that R2 and R3 are in a Parallel combination,hence thier net resistance will be given by=>

1/Rp = 1/R2 + 1/R3

=>1/Rp = 1/8 + 1/8

=>1/Rp = 2/8

=>Rp = 4ohm.

hence the net resistance of the Parallel combination is 4 ohm.

ii)Now, since Rp and R1 are in series combination,hence their net resistance is given by-->>

Rn = Rp + R1

=>Rn = 4 + 4

=>Rn = 8ohm.

hence the net resistance of the whole circuit is 8 ohm.

Now, by using ohm 's law >

V= IR

=>I = V/R

=> I = 8/8

=>I = 1 ampere.

So , the current flowing through 5ohm resistor is 1 ampere.

iii)Now, again by Ohm's law>>

V=IR

=>V = 1 × 4

=>V = 4 v.

hence the potential difffernce across 4 ohm resistor is 4volts.

iv)We know that,

Power used = V × I

=>P = 4 × 1

=>P = 4 Watts.

Hence ,the power dissipated in 4 ohm resistor is 4 watts.

v) Same reading of both ammeters.

Since we know that the resistances Rp and R1 are in series combination hence same Current will flow through each resistor.Hence ammeter reading will be same.

Remember

1)Series Combination

-Net resistance is sum of all resistance in series.

-Rn= R1 + R2 + R3 +..... Rg

where,

Rn=> is the net resistance

R1,R2,R3=> are resistors in series and

Rg =>is the last resistor of the circuit.

-In these type of circuits,the net resistance is always greater than the highest value of resistance in the circuit.

2) Parallel Combination

- in this type of connection inverse of Net resistance is the sum of inverse of all other resistance in the circuit.

-1/Rn = 1/R1 + 1/R2 + 1/R3 +....1/Rp

where,

1/Rn=> is inverse for net resistance

1/Rp is the inverse of last resistance in the circuit.

-In this type of combination the net resistance is smaller than the smallest value of resistance in the combination.

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Answer:

Explanation:

Given :-

Resistor, R₁ = 4 Ω

Resistor, R₂ = 8 Ω

Resistor, R₃ = 8 Ω

Potential Difference, V = 8 volts.

Solution :-

1. Effective resistance of two 8 Ω resistors in the combination.

Solution 1.

The two 8 Ω resistances are connected in parallels. If R is the effective resistance of these, Then

1/R = 1/8 Ω + 1/8 Ω

1/R = (1 + 1)/8 Ω

1/R = 1/4 Ω

R = 4 Ω

Hence, 4 Ω is effective resistance of two 8 Ω resistors in the combination.

2. Current flowing through 4 Ω resistor.

Solution 2.

The parallel combination of 8 ohm Ω resistances forms a series combination of 4 Ω resistance with the other 4 Ω resistance.

Then,

Total resistance in the Circuit,    R = 4 Ω + 4 Ω = 8 Ω

Applied Voltage,                          V = 8 volts

So Current Flowing,                     I = Voltage/Resistance

                                                      I = 8 V/8 Ω

                                                     I = 1 Ampere.

Hence, the current flowing through 4 Ω resistor is 1 ampere.

3. Potential difference across 4 Ω resistor.

Solution 3.

Potential Difference across 4 Ω resistance = Current × Resistance

                                                                             = 1 A × 4 Ω

                                                                             = 4 volts.

Hence, the Potential difference across 4 Ω resistor is 4 volts.

4. Power dissipated by  4 Ω resistor.

Solution 4.

Power dissipated in 4 Ω Resistor = 1 V = 1 A × 4 V = 4w

Explanation: