Question

a) Which two resistors are connected in series ?
b) Which two resistors are connected in parallel ?
c) If every resistor of the circuit is of 2 2, what current will flow in the circuit ?

Answer 1

Answer:

a) The resistances $R_{5}$ and $R_{4}$ are connected in series.

b) The resistances $R_{2}$ and $R_{3}$ are connected in parallel.

c) The current in the circuit is 1A.

Explanation:

For part a,

The resistances $R_{5}$ and $R_{4}$ are connected in series.

For part b,

The resistances $R_{2}$ and $R_{3}$ are connected in parallel.

For part c,

Given:

$R_{1}=R_{2} =R_{3 =R_{4} =R_{5}= 2ohm$

$v=5V$

Here,

The equivalent resistance of the resistances $R_{2}$  and $R_{3}$ is denoted by $R_{23}$.

The equivalent resistance in the circuit is denoted by $R_{eq}$.

The current in the circuit is denoted by I.

The emf of the battery is denoted by V.

Now,

By the equation,

$\frac{1}{R_{23} } =\frac{1}{R_{2} } +\frac{1}{R_{3} } \\\frac{1}R_{23} } =\frac{1}{2}+ \frac{1}{2} \\R_{23} =1ohm$

Then,

By the equation,

$R_{eq} =R_{23} +R_{5} +R_{4} \\R_eq}=1+2+2 \\R_{eq} =5ohm$

Now,

By the equation,

$I=\frac{V}{R_{eq} } \\I=\frac{5}{5}\\ I=1A$

So, the current in the circuit is 1A.

Answer 2

Answer:

→ I = 1 ampere (ans)

Explanation:

a) Resistors R4 and R5 are in series.

b)  Resistors R2 and R3 are in parallel.

c) Current flowing from circuit is 1 ampere.

Step-be-step explanation:

As per the given data from the image,

a) We can clearly see that resistors R4 and R5 are connected In series,

So they add up directly.

b) We can clearly see that resistors R2 and R3 are connected in parallel

As they have common point at both the ends of them.

c) If every resistor of the circuit is of 2 Ω, then the current that will flow from the circuit is:

First we will calculate the equivalent resistance of resistors connected in parallel which are R2 and R3

From the formula of equivalent resistance (Req) in parallel circuit

1/Req = 1/R2 + 1/R3

Now,

Substituting R2  and R3 = 2 Ω in the formula

1/Req = ½ + ½

→ 1/Req = 1

Thus Req = 1 Ω

Now as we can clearly see that resistor R1 is short circuited  by a plain wire

So it will not be taken into consideration

Now Req  of series circuit is

Substituting values of all the resistances we get

Req = R5 + Req(of parallel circuit) + R4

Req = 2 + 1 + 2 = 5 Ω

Now as we know that

V = IR

Here V = 5 volts

I is to be calculated and R (resistance of the circuit) is 5 Ω

Substituting values in the formula we get,

5 = I×5

→ I = 1 ampere (ans)