Question

Physics question please solve please don't give unnecessary answers give answers for all parts please it's urgent​

Answer 1

Answer with explanation :

$\bf \: \bigstar \: a \\ \\ \sf \: here \: \: 4\Omega \: \: and \: 4\Omega \: \: resistor \: are \: connected \\ \sf\: in \: parallel \\ \sf \therefore \: \frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \\ \sf \therefore \: R_p = 2\Omega \\ \\ \sf \: again \: \: R_p \: \: and \: \: 2\Omega \: \: are \: in \: series \: \\ \sf \therefore \: eqiuvalent \: resistance = R_{eq} \\ \\ R_{eq}= R_p + 2 = 2 + 2 = 4\Omega \: \\ \sf \therefore \: R_{eq} = 4\Omega \\ \\ \\ \pink { \boxed{ \sf \: equivalent \: resistance \: \: R_{eq} = 4\Omega}}$

$\bigstar \: \bf \: b \\ \sf \: ammeter \: reading \: = I = \: to \: find \\ \\ \bf \: given \: \\ \sf \: voltage \: \: V = 4 \: V \\ \sf \: resistance \: of \: the \: circuit \: \: R_{eq} = 4\Omega \: \\ \\ \bf \: we \: know \: that \: \\ \sf \: V = IR_{eq} \\ \sf \therefore \: I = \frac{V}{R_{eq}} \\ \sf \: \: \: \: \: \: \: = \frac{4}{4} = 1 \: A \\ \\ \green { \boxed{ \sf \: ammeter \: reading \: I = 1 \: A}}$

$\bigstar \bf \: c \\ \\ \sf \: voltmeter \: reading \: \: V_m = IR_p \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 1 \times 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2V \\ \\ \sf \orange{ \boxed{ \sf \: voltmeter \: reading \: V_m = 2 \: V}}$

$\bf \bigstar \: d$

We know that,

The ammeter has to be connected in series to the circuit. As a result, if we add it in parallel, its reading will appear as zero.

In a same way,

The voltmeter has to be connected in parallel the circuit. As a result, if we add it in series , its reading will appear as zero.

Answer 2

Answer:

Answer :-

$\bf \: \bigstar \: a \\ \\ \sf \: here \: \: 4\Omega \: \: and \: 4\Omega \: \: resistor \: are \: connected \\ \sf\: in \: parallel \\ \sf \therefore \: \frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \\ \sf \therefore \: R_p = 2\Omega \\ \\ \sf \: again \: \: R_p \: \: and \: \: 2\Omega \: \: are \: in \: series \: \\ \sf \therefore \: eqiuvalent \: resistance = R_{eq} \\ \\ R_{eq}= R_p + 2 = 2 + 2 = 4\Omega \: \\ \sf \therefore \: R_{eq} = 4\Omega \\ \\ \\ \pink { \boxed{ \sf \: equivalent \: resistance \: \: R_{eq} = 4\Omega}}$

$\bigstar \: \bf \: b \\ \sf \: ammeter \: reading \: = I = \: to \: find \\ \\ \bf \: given \: \\ \sf \: voltage \: \: V = 4 \: V \\ \sf \: resistance \: of \: the \: circuit \: \: R_{eq} = 4\Omega \: \\ \\ \bf \: we \: know \: that \: \\ \sf \: V = IR_{eq} \\ \sf \therefore \: I = \frac{V}{R_{eq}} \\ \sf \: \: \: \: \: \: \: = \frac{4}{4} = 1 \: A \\ \\ \green { \boxed{ \sf \: ammeter \: reading \: I = 1 \: A}}$

$\bigstar \bf \: c \\ \\ \sf \: voltmeter \: reading \: \: V_m = IR_p \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 1 \times 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2V \\ \\ \sf \orange{ \boxed{ \sf \: voltmeter \: reading \: V_m = 2 \: V}}$

$\bf \bigstar \: d$

We know that,

The ammeter has to be connected in series to the circuit. As a result, if we add it in parallel, its reading will appear as zero.

In a same way,

The voltmeter has to be connected in parallel the circuit. As a result, if we add it in series , its reading will appear as zero.