Explanation:
Given that,
Two resistors 4.5 Ω and 2.3 Ω .
Potential difference = 30 V
When they are in series, the current through each resistor remains the same. First find the equivalent resistance.
R' = 4.5 + 2.3
= 6.8 Ω
Current,
[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]
So, the current through both lightbulb is the same i.e. 4.41 A.
When they are in parallel, the current divides.
Current flowing in 4.5 resistor,
[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]
Current flowing in 2.3 ohm resistor,
[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]
In parallel combination, are brighter than bulbs in series.
