Question

Find out the following in the electric circuit given in Figure

1. Effective resistance of two 8 Ω resistors in the combination
2. Current flowing through 4 Ω resistor
3. Potential difference across 4 Ω resistance
4. Power dissipated in 4 Ω resistor
5. Difference in ammeter readings, if any.

Answer 1

1. Both 8Ω resistors are connected in parallel so the effective resistance of the two will be 
   `1/"R"_"P" = 1/"R"_1 + 1/"R"_2`
   `1/"R"_"P" = 1/8 + 1/8`
   `1/"R"_"P" = 2/8`
   `1/"R"_"P" = 1/4`
   R_p = 4Ω
2. Net resistance of the circuit will be 
   R_net = 4 + 4
   R_net = 8Ω
   So, current in 4Ω resistor using Ohm's law is
   `"I" = "V"/"R"_"net"`
   `"I" = 8/8`
   I = 1 A
3. Potential difference across 4Ω resistor is
   V = IR
   V = 1 × 4
   V = 4 V
4. Power dissipated through 4Ω is
   P = i²R
   P = 1² × 4
   P = 4 W
5. Both ammeters A₁ and A₂ will record same readings because they are connected in series. Hence difference in their readings will be zero.