Physics Current Electricity

Using Heat equation : H = i²Rt

=>192 = (4)² R (1)

H = (8)² R (5)

=>H = 3840 J

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

Since $5\Omega$ is connected across conductor so we can remove it.

$R_{eq}=1\Omega$

The basic unit of energy is a Joule (J), which is equal to one watt of power expended for one second. In the day-to-day scenarios such as household electricity consumption, joule is too small a unit to be convenient. For billing and metering purpose, it is measured in kilowatt-hours (kWh). Electrical energy is measured in kilowatt-hours (kWh) as the commercial unit. 1kWh = 1000 W * 3600 s 

For calculating the units consumed, we will be using the following formula 

Units = [Power (W) x Time (h)]/1000

The SI unit of the electrical power is Watt which is symbolized as W. One watt is equal to one joule per second (J/s). In the context of current and voltage, one watt is the power consumed when one ampere of the current flows through the potential difference of one volt.

$R_{eq}=3\Omega$

When switch is open,

$R_{eq}=\frac{3R}{2}$              

When switch is closed,

$R_{eq}^{\text{'}}=2*\frac{R*2R}{R+2R}=\frac{4R}{3}$           

$\frac{R_{eq}}{R_{eq}^{\text{'}}}=\frac{3R/2}{4R/3}=\frac{9}{8}=\frac{x}{8}$

$\Rightarrow x=9$

In series Req = nR = 10n

$\Rightarrow ls=\frac{20}{10+10n}=\frac{2}{1+n}$

In parallel $\Rightarrow R_{eq}=\frac{10}{n}$ ${}_{}\frac{}{}$

$\Rightarrow \frac{l_{\rho }}{l_s}=20=\frac{2n/1+n}{2/1+n}$

n = 20

Given

$I=\frac{2E}{R=r_1+r_2}$

E₁ = E₂ = E - (i)

Potential drop across second cell is

$V_A-E_2+I{r}_2=v_B$

According to question VA – VB = 0

E2 lr2 = 0

$\Rightarrow R=r_2-r_1$