2 resistors of resistance 1000 ohm and 2000 ohm are joined in series with a 100V supply. A voltmeter of internal resistance 4000 ohm is conn

Question

2 resistors of resistance 1000 ohm and 2000 ohm are joined in series with a 100V supply. A voltmeter of internal resistance 4000 ohm is connected to measure the potential difference across 1000 ohm resistor. Calculate the reading shown by the voltmeter.

in progress 0
RI SƠ 3 years 2021-08-20T21:51:48+00:00 1 Answers 252 views 0

Answers ( )

    0
    2021-08-20T21:52:59+00:00

    The voltmeter reading will be 35.7 volt

    Explanation:

    The resistor 1000 ohm and 4000 ohm are connected in parallel .

    Their combined resistance is supposed R₁

    Thus \frac{1}{R_1} = \frac{1}{1000} + \frac{1}{4000}  

    or R₁ = 800 ohm

    Therefore the total resistance in circuit is = 2000 + 800 = 2800 ohm

    Because these are in series .

    We can find  current flowing through the circuit  I = \frac{V}{R} = \frac{100}{2800} = \frac{1}{28}

    here R is total resistance .

    The potential difference across 1000 ohm = \frac{1}{28} x 1000 = 35.7 volt

    Thus voltmeter reading will be 35.7 volt

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )