An electric lamp of resistance $20 Ω$ and a resistor of resistance $4 Ω$ are connected to a 6 V battery as shown in the circuit diagram.
What is the value of
A. The total resistance of the circuit and
B. The current through the circuit.
C. The potential difference across the
1. electric lamp
2. Resistor
D. Power of the lamp.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

(A)25ohm
(B)0.50A
(C)(1)6V (2)2V
(D)0.25W

(A)24ohm
(B)0.25A
(C)(1)5V (2)1V
(D)1.25W

(A)20ohm
(B)0.20A
(C)(1)4V (2)0.5V
(D)1.35W

(A)22ohm
(B)0.30A
(C)(1)4.5V (2)2.5V
(D)0.25W

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given:
electric resistance is $R_{1} = 20 Ω$
Resistance is $R_{2} = 4 Ω$
power deliver, V = 6 V.
A. the absolute resistance of the circuit is,
$R_{T} = R_{1} + R_{2}$
The rate adjustments as,
$R_{T} = 24 Ω$
therefore, the maximum resistance cost is 24Ω
.
B. The circuit breaker says,
$I_{C} = V / R_{T}$
The fee modifications as,
$I_{C} = 6 V / 24 Ω$

$I C = 0.25 A$
consequently, the current rotation limit is zero.25 A.
C. 1. The power of the electric lamp that,
$V_{1} = I_{C} \times R_{1}$
The price changes as,
$V_{1} = 0.25 A \times 20 Ω$

$V_{1} = 5 V$
therefore, the most voltage throughout the electric lamp is five V.
C.2 power of each driver,
$V_{2} = I_{C} \times R_{2}$
The rate adjustments as,
$V_{2} = 0.25 A \times 4 Ω$

$V_{2} = 1 V$
consequently, the voltage range across the conductor is 1 V.
D. The electricity of the lamp may be calculated as,
P = V1 × IC
The price adjustments as,
P = 5V × 0.25A

P = 1.25W
consequently, the maximum power according to lamp is 1.25W
.

Watch 3-min video & get full concept clarity