Question

Calculate the total resistance and current in the circuit if R2,R3,R4 are 2 ohm and R1 is 4 ohm.potential difference across the battery is 12 V.

Answer 1

Answer:

total resistance 0.75 ohm

total current 16 ampere

Explanation:

as R1 and R4 are parallel to each other

=1/2+1/4

=3/4

=0.75

current = voltage ÷ resistance

current = 12/0.75

current = 16

Answer 2

Answer:

The total resistance, R, measured is $2.4\Omega$.

The current flow in the circuit, I, measured is $5A$.

Explanation:

Data given,

The value of the resistances, R₂, R₃, R₄ = $2\Omega$

The value of the resistance R₁ = $4\Omega$

The potential difference across the given battery, V = $12V$

The total resistance, R =?

The current flow, I =?

From the given figure, we can see that the resistances, R₂, R₃, R₄ are all connected in series.

Thus,

• Resistance in series, $R_{S}$ = $R_{2} + R_{3} +R_{4}$ = $2 + 2 + 2$ = $6\Omega$

Also, from the figure, we can see that the resistances, R₂, R₃, R₄ ( $R_{S}$ ) are connected in parallel with R₁.

Thus,

• Resistance in parallel, $\frac{1}{R}$ = $\frac{1}{R_{1} } + \frac{1}{R_{S} }$ = $\frac{1}{4} + \frac{1}{6}$ = $\frac{6 +4}{24}$ = $\frac{10}{24}$

Now,

• Total resistance, R = $\frac{24}{10}$ = $2.4\Omega$

Now, we can find out the current by the equation, $V = IR$

• $I = \frac{V}{R}$ = $\frac{12}{2.4}$ = $5A$

Hence, the current flow in the circuit, I, measured is $5A$.