Question

A combined circuit has two resistors in parallel (6.0 ohms and 8.0 ohms) and another in series (13.0 ohms). If the power source is 9.0 volts, what will the current be?

Answer 1

Answer:

The current is 0.548 Amperes

Explanation:

• When resistors are arranged in series to get total resistance we add the resistance of resistors.

That is; Rt = R₁ +R₂ +R₃ + ...........+ Rn

• For resistors in parallel the effective resistance is given by;

$\frac{1}{Rt} =\frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} + .....+\frac{1}{Rn}$

Step 1: Effective resistance for resistors in parallel

$\frac{1}{Rt} =\frac{1}{R1} +\frac{1}{R2}$

$\frac{1}{Rt} =\frac{1}{6} +\frac{1}{8}$

$\frac{1}{Rt} =\frac{7}{24}$

$Rt=3.43 ohms$

Step 2: Total resistance for the circuit

Resistors 3.43 ohms and 13 ohms are in series

For resistors in series.

Rt = R₁ +R₂

Thus, effective resistance = 3.43 ohms + 13 ohms

                                           = 16.43 ohms

The total resistance for the circuit is 16.43 ohms

step 3: Current in the circuit

Total resistance = 16.43 ohms

Voltage = 9.0 volts

But; from the Ohm's law, I =V/R

Thus, current, $I = \frac{9}{16.43} \\= 0.548 Amperes$

Answer 2

Answer:

The current will be 0,54 amperes

Explanation:

The circuit has the  appearance from the scheme.

6 Ω and 8 Ω are the resistances in parallel, so to calculate the equivalent resistance we have to add the inverses of each one and then raise it to -1

(Req)*-1 = 1/6 + 1/8 = 24/7 Ω

24/7 Ω + 13 Ω (both are in series, we have to sum each other) = 115/7 Ω

THIS IS THE TOTAL RESISTANCE.

Now we can apply The Ohm's law.

ΔV = i . R

9 V / 115/7 Ω = 0.54 Amperes