Question

PHYSICS II:

4) For the circuit shown below, find: a) the equivalent resistance; b) the current through the battery; c) the potential difference across the 10 ohm resistor. 9 p /A we 100C 5) Two resistors with resistances of 12.0 2 and 6.0 are connected in parallel. An 8.0-12 resistor is then connected in series with this parallel combination. An ideal 12.0-V battery is then connected across the series- parallel combination. A) Draw a diagram of this circuit. B) Find the equivalent resistance of the circuit.

Answer 1

(5)

(A) circuit diagram:- 1282 ww GR - Equivalent Resistance of 12 x 862 = t + 5 = 112 R=42 This is in paralled a series with 8r. 80, Equivalent Resistance of circuit I 4+ 8 = 122

(4) the two resistors are in series so

(a) the equivalent resistance R_{​​​​​​eq}​​​​ = 10+ $\ohm$ 5 = 15 ohm

(b) current = V/R = 9/15 = 0.6 A

(c) voltage through 10 ohm resistor = IR = 10× 0.6 = 6 V