Question

The figure given below shows an electric circuit in which current flows from a 6V battery through two resistors.; (a) Are the resistors connected in series with each other or in parallel?; (b) For each resistor, state the p.d. across it.; (c) The current flowing from the battery is shared between the two resistors. Which resistor will have bigger share of the current; (d) Calculate the effective resistance of the two resistors.; (e) Calculate the current that flows from the battery.

Answer 1

The figure given below shows an electric circuit in which current flows from a 6V battery through two resistors.;

• (a) Are the resistors connected in series with each other or in parallel?;
• The registers are connected in parallel
• (b) For each resistor, state the p.d. across it.;
• Since, the resistors are in parallel, the p.d across each resistors remains the same, i.e, 6V
• (c) The current flowing from the battery is shared between the two resistors. Which resistor will have bigger share of the current;
• we have, R=V/I
• the resistor with smaller amount of resistance will have bigger share of current. i.e, 2 ohm resistor.
• (d) Calculate the effective resistance of the two resistors.;
• $\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\=\dfrac{1}{2}+\dfrac{1}{3}\\\\=\dfrac{3+2}{3 \times 2}\\\\=\dfrac{5}{6}\\\\R=\dfrac{6}{5}$
• ∴R=1.2 ohm.
• (e) Calculate the current that flows from the battery.
• we have,
• I = V/R
• =6/1.2
• ∴I= 5A

Answer 2

The image given in question is attached below.

(a) The resistor are connected in parallel connection.

(b) Potential difference across each resistor is 6 V:

• When resistors are attached in parallel connection, the voltage remains same in each resistor.
• From circuit, we can see that the battery is connected parallel to the connection.
• So, the potential difference across each resistor is 6 V.

(c) 2 Ω resistor have big share:

• The current flowing through the wire is inversely proportional to the resistance of the wire.
• The 3 Ω resistor shows more resistance then 2 Ω resistor.
• Thus, resistor with less resistance have bigger share of current.

(d) Effective resistance of the two resistors is 1.2 Ω

1/R = 1/R₁ + 1/R₂

1/R = 1/2 + 1/3

1/R = (3+2)/6

1/R = 5/6

R = 6/5

∴ R = 1.2 Ω

(e) Current flowing through wire is 5 ampere.

V = IR ⇒ I = V/R

On substituting the values, we get,

∴ I = 6/1.2 = 5 A