Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.

(a) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in series with a battery of 6.0 V and negligible internal resistance is shown below:

Given,

Two resistors of 4 Ω and 6 Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (4 + 6) Ω = 10 Ω

Potential Difference V = 6 V

Current I = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

6 = I x 10
⇒ I = 6 / 10 = 0.6 A

Hence, in series, current through the battery = 0.6 A

(b) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in parallel with a battery of 6.0 V and negligible internal resistance is shown below:

Given,

Two resistors of 4 Ω and 6 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R$p} = \dfrac{1}{4} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{3 + 2}{12} = \dfrac{5}{12} \\[0.5em] R_p = \dfrac{12}{5} = 2.4 Ω

Potential Difference V = 6 V

Current I = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

6 = I x 2.4 ⇒ I = 6 / 2.4 = 2.5 A

Hence, in parallel, current through the battery = 2.5 A