Question

A block of metal is heated directly by dissipating power in the internal resistance of block. Because of temperature rise, the resistance increases exponentially with time and is given by R(t) = 0.5 e^2t, where t is in second. The block is connected across a 110 V source and dissipates 7644 J heat energy over a certain period of time. This period of time is ______ × 10^-1 sec (take ln 0.367 = -1).

Options

• 5

• 6

• 7

• 8

Answer 1

A block of metal is heated directly by dissipating power in the internal resistance of block. Because of temperature rise, the resistance increases exponentially with time and is given by R(t) = 0.5 e^2t, where t is in second. The block is connected across a 110 V source and dissipates 7644 J heat energy over a certain period of time. This period of time is 5 × 10^-1 sec (take ln 0.367 = -1).

Explanation:

Let t be the required time. As power is

P = `"dU"/"dt" = "V"^2/("R"("t"))`

dU = `"V"^2/("R"("t"))`dt

U = `int_0^t"V"^2/("R"("t"))`dt

= `(110)^2/0.5int_0^"t" "e"^(-2"t") "dt"`

= `(110)^2/(2xx0.5) ("e"^(-2"t"))_0^"t"`

= (110)² (1 - e^-2t)J 

According to problem,

U = 7644 J

Thus  1 - e^-2t = `7644/(110)^2` = 0.632

or e^–2t = 0.367

or –2t lne = ln 0.367

or –2t = –1

t = 5 × 10^–1 sec