Question

Combine circle has two resistors in parallel (34.0 ohms and 41.0 ohms) and another in series (15.0 ohms) if the power source is 9.0 volts what will the current be?

Answer 1

-- If you have 34 ohms and 41 ohms in parallel,
they behave like a single resistor of  18.59 ohms.

-- With another 15 ohms in series with those,
the battery sees a total resistance of  33.59 ohms.

-- Current = (voltage) / (resistance)

Current = (9 v) / (33.59 ohms)

Current = 0.268 Ampere

Answer 2

Answer:

The current will be 0.27 amps

Explanation:

Electric resistance is defined as any opposition that encounters the current passing through a closed electrical circuit, attenuating the free flow of circulation of electric charges or electrons.

Resistance or equivalent resistance is called the value of the resistance obtained by associating a set of them.

In a parallel resistor circuit the current is divided and circulates in several ways. Then the total equivalent resistance is equal to the reciprocal of the sum of the inverse of the individual resistances.

For this case:

$Reqp(total parallel resistance)=\frac{1}{\frac{1}{R1}+\frac{1}{R2} }$

$Reqp(total parallel resistance)=\frac{1}{\frac{1}{34}+\frac{1}{41} }$

Reqp (total parallel resistance) = 18.59 ohms

On the other hand, the resistors are in series when they are connected next to each other, that is, the output point of a resistor is the entry point of the resistor. The resistance value equivalent to the resistors connected in series is equal to the sum of the values ​​of each of them. In this case

In this case the series combination is generated between parallel Reqp calculated previously and R = 15.0  ohms. Then

Reqs=Reqp+15.0 ohms=18.59 ohms+15 ohms= 33.59 ohms

Ohm's Law indicates that current intensity that passes through a circuit is directly proportional to its voltage or voltage and inversely proportional to the resistance it presents:

$I=\frac{V}{R}$

where I is the intensity that is measured in amps (A), V the voltage that is measured in volts (V); and R the resistance measured in ohms (Ω).

In this case R=Reqs and V= 9.0 volts

Then:  

$I=\frac{9 volts}{33.59 ohms}$

I= 0.27 amps

The current will be 0.27 amps