Question

please answer the following questions​

Answer 1

Answer:

=R1+(R2+R3)

= 4+(1/3+1/6)

=4+0.5

= 4.5ohm

I=V/R

= 12/4.5

=8/3Ampere

Answer 2

Given :

• R₁ = 4 ohms
• R₂ = 3 ohms
• R₃ = 6 ohms

potential difference = 12 volts

To find :

The total resistance and the total current flowing through it.

Solution :

According to the question,

R₃ and R₂ are connected in parallel combination.

we know that,

The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

1/R = 1/R₁ + 1/R₂

By substituting the values in the formula,

➛ 1/R₂₃ = 1/R₂ + 1/R₃

➛ 1/R₂₃ = 1/3 + 1/6

➛ 1/R₂₃ = 2+1/6

➛ 1/R₂₃ = 3/6

➛ R₂₃ = 2 ohms

Now, R₂₃ and R₁ are connected in series combination,

» The combined resistance of any number of resistance connected in series is equal to the sum of the individual resistances. i.e.,

R = R₁ + R₂

By substituting the values in the formula,

➛ R = R₁ + R₂₃

➛ R = 4 + 2

➛ R = 6 ohms

Thus, the total resistance is 6 ohms.

To find the current we can use ohm's law that is,

» At constant temperature the current flowing through a conductor is directly proportional to the potential difference across its ends.

Formula : V = RI

where,

• V denotes potential difference
• R denotes resistance
• I denotes current

substituting all the given values in the formula,

➛ V = RI

➛ I = V/R

➛ I = 12/6

➛ I = 2 A

Thus, the total current of the system is 2 Ampere.

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