Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
to find i1, i2, and i3 we need to find the total current.
to find the total current, you need to find the total resistance
you're already given the total voltage, Vs
to find Rtotal, start from the resistors furthest from the voltage source.
R3 and R4 are in series so
Rtotal= R3+R4 = 6+3 = 9 ohms
9 ohms is now in parallel with R2 so,
Rtotal= + (\frac{1}{R2}) ^{-1})^-1= (1/18)^-1 +( 1/9)^-1 = 6 ohms
6 ohms is in series with R1 so
Rtotal= 4+6=10 ohms
itotal= (\frac{Vtotal}{Rtotal})
= 120 v/10 ohms = 12 A
i total = i1 because all the current flows through it
i1= 12A
so the current splits into i2 and i3 and the amount of current that flows through a branch depends on the total resistance in each branch.
we already calculated the resistance in the R3+R4 & R2 branch as 6 ohms
since r3 and r4 are in series, the same current will flow through them
r3+r4 = 9 ohms
r2= 18 ohms
so the current in r2 will be half that of R3 & R4 (V=IR)
using the current divider rule
i2= 12A x (6 ohms/18 ohms)= 4 A
i3= 12A x (6 ohms/9 ohms) = 8 ohms