Đáp án:

$I\approx 5,27A$

$I_{đ1}\approx 0,45A$

$I_{đ2}\approx 0,27A$

$I_{b}\approx 4,5A$

Giải thích các bước giải:

 $R_{đ1}=\frac{U_{đ1}^2}{P_{đ1}}=\frac{220^2}{100}=484\Omega$

 $R_{đ2}=\frac{U_{đ2}^2}{P_{đ2}}=\frac{220^2}{60}\approx 806,6\Omega$

 $R_{b}=\frac{U_{b}^2}{P_{b}}=\frac{220^2}{1000}=48,4\Omega$
$R_{tđ}=\frac{1}{\frac{1}{R_{đ1}}+\frac{1}{R_{đ2}}+\frac{1}{R_b}}\\=\frac{1}{\frac{1}{484}+\frac{1}{806,6}+\frac{1}{48,4}}\approx 41,72\Omega$

a, $I=\frac{220}{R_{tđ}}\approx 5,27A$

$I_{đ1}=\frac{P_{đ1}}{U_{đ1}}=\frac{100}{220}\approx 0,45A$

$I_{đ2}=\frac{P_{đ2}}{U_{đ2}}=\frac{60}{220}\approx 0,27A$

$I_{b}=\frac{P_{b}}{U_{b}}=\frac{1000}{220}\approx 4,5A$

b,

Công suất tiêu thụ:

$E=P.t=(100+60).4.30+1000.5.30=169200Wh=169,2kWh$

Số tiền:

$M=169,2.1300=219960vnđ$